Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Integration by Partial Fraction
Decomposition
Bernd Schröder
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Introduction
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Introduction
1. Polynomials p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 are
easy to integrate.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Introduction
1. Polynomials p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 are
easy to integrate.
2. Rational functions are functions of the form
an xn + an−1 xn−1 + · · · + a1 x + a0
f (x) =
bm xm + bm−1 xm−1 + · · · + b1 x + b0
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Introduction
1. Polynomials p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 are
easy to integrate.
2. Rational functions are functions of the form
an xn + an−1 xn−1 + · · · + a1 x + a0
f (x) =
bm xm + bm−1 xm−1 + · · · + b1 x + b0
3. They are a natural generalization of polynomials
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Introduction
1. Polynomials p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 are
easy to integrate.
2. Rational functions are functions of the form
an xn + an−1 xn−1 + · · · + a1 x + a0
f (x) =
bm xm + bm−1 xm−1 + · · · + b1 x + b0
3. They are a natural generalization of polynomials, and they
occur pretty frequently in applications, too.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Introduction
1. Polynomials p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 are
easy to integrate.
2. Rational functions are functions of the form
an xn + an−1 xn−1 + · · · + a1 x + a0
f (x) =
bm xm + bm−1 xm−1 + · · · + b1 x + b0
3. They are a natural generalization of polynomials, and they
occur pretty frequently in applications, too.
4. So we are interested in the integrals of rational functions.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Introduction
1. Polynomials p(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 are
easy to integrate.
2. Rational functions are functions of the form
an xn + an−1 xn−1 + · · · + a1 x + a0
f (x) =
bm xm + bm−1 xm−1 + · · · + b1 x + b0
3. They are a natural generalization of polynomials, and they
occur pretty frequently in applications, too.
4. So we are interested in the integrals of rational functions.
5. The method we use, partial fraction decomposition, is also
very important for solving differential equations with
Laplace transforms.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Theorem.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Theorem. Real number version of the Fundamental
Theorem of Algebra.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Theorem. Real number version of the Fundamental
Theorem of Algebra. Every polynomial with real coefficients
n
can be factored
ninto factors of the form (x − c) and
2
2
(x − a) + b .
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Theorem.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Theorem. Partial Fraction Decomposition.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Theorem. Partial Fraction Decomposition. Every rational
function with real coefficients
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Theorem. Partial Fraction Decomposition. Every rational
function with real coefficients so that the degree of the
numerator is smaller than that of the denominator
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Theorem. Partial Fraction Decomposition. Every rational
function with real coefficients so that the degree of the
numerator is smaller than that of the denominator can be
written as a sum of terms of the form
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Theorem. Partial Fraction Decomposition. Every rational
function with real coefficients so that the degree of the
numerator is smaller than that of the denominator can be
A
written as a sum of terms of the form
(x − c)n
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Theorem. Partial Fraction Decomposition. Every rational
function with real coefficients so that the degree of the
numerator is smaller than that of the denominator can be
A
and
written as a sum of terms of the form
(x − c)n
Ax + B
n.
((x − c)2 + b2 )
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Theorem. Partial Fraction Decomposition. Every rational
function with real coefficients so that the degree of the
numerator is smaller than that of the denominator can be
A
and
written as a sum of terms of the form
(x − c)n
Ax + B
n . In each case, n runs from 1 to the exponent
((x − c)2 + b2 )
that the respective denominator has in the factorization of the
function’s denominator.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Theorem. Partial Fraction Decomposition. Every rational
function with real coefficients so that the degree of the
numerator is smaller than that of the denominator can be
A
and
written as a sum of terms of the form
(x − c)n
Ax + B
n . In each case, n runs from 1 to the exponent
((x − c)2 + b2 )
that the respective denominator has in the factorization of the
function’s denominator.
So if we can integrate these functions and if we can find a way
to write out the decomposition, then we can integrate rational
functions.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Proposition.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Proposition. Integrals of the rational functions that cannot be
further decomposed.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Proposition. Integrals of the rational functions that cannot be
further decomposed. (Integration constants omitted.)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Proposition. Integrals of the rational functions that cannot be
further
Z decomposed. (Integration constants omitted.)
1
1.
dx = ln |x − c|
x−c
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Proposition. Integrals of the rational functions that cannot be
further
Z decomposed. (Integration constants omitted.)
1
1.
dx = ln |x − c|
Z x−c
1
1
2.
dx = −
n
(x − c)
(n − 1)(x − c)n−1
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Proposition. Integrals of the rational functions that cannot be
further
Z decomposed. (Integration constants omitted.)
1
1.
dx = ln |x − c|
Z x−c
1
1
(n > 1)
2.
dx = −
n
(x − c)
(n − 1)(x − c)n−1
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Proposition. Integrals of the rational functions that cannot be
further
Z decomposed. (Integration constants omitted.)
1
1.
dx = ln |x − c|
Z x−c
1
1
(n > 1)
2.
dx = −
n
(n − 1)(x − c)n−1
Z (x − c)
1
1
x
3.
dx = arctan
2
2
b
b
x +b
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Proposition. Integrals of the rational functions that cannot be
further
Z decomposed. (Integration constants omitted.)
1
1.
dx = ln |x − c|
Z x−c
1
1
(n > 1)
2.
dx = −
n
(n − 1)(x − c)n−1
Z (x − c)
1
1
x
3.
dx = arctan
2
2
b
b
Z x +b
1
x
+
4.
dx
=
n
2 (x2 + b2 )n−1
(x2 + b2 )
2(n
−
1)b
Z
2n − 3
1
+
dx
2
n−1
2(n − 1)b
(x2 + b2 )
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Proposition. Integrals of the rational functions that cannot be
further
Z decomposed. (Integration constants omitted.)
1
1.
dx = ln |x − c|
Z x−c
1
1
(n > 1)
2.
dx = −
n
(n − 1)(x − c)n−1
Z (x − c)
1
1
x
3.
dx = arctan
2
2
b
b
Z x +b
1
x
+
4.
dx
=
n
2 (x2 + b2 )n−1
(x2 + b2 )
2(n
−
1)b
Z
2n − 3
1
+
dx
(n > 1)
2
n−1
2(n − 1)b
(x2 + b2 )
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Proposition. Integrals of the rational functions that cannot be
further
Z decomposed. (Integration constants omitted.)
1
1.
dx = ln |x − c|
Z x−c
1
1
(n > 1)
2.
dx = −
n
(n − 1)(x − c)n−1
Z (x − c)
1
1
x
3.
dx = arctan
2
2
b
b
Z x +b
1
x
+
4.
dx
=
n
2 (x2 + b2 )n−1
(x2 + b2 )
2(n
−
1)b
Z
2n − 3
1
+
dx
(n > 1)
2
2 + b2 )n−1
2(n − 1)b
(x
Z
x
1 2
2
5.
dx
=
ln
x
+
b
2
x2 + b2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Proposition. Integrals of the rational functions that cannot be
further
Z decomposed. (Integration constants omitted.)
1
1.
dx = ln |x − c|
Z x−c
1
1
(n > 1)
2.
dx = −
n
(n − 1)(x − c)n−1
Z (x − c)
1
1
x
3.
dx = arctan
2
2
b
b
Z x +b
1
x
+
4.
dx
=
n
2 (x2 + b2 )n−1
(x2 + b2 )
2(n
−
1)b
Z
2n − 3
1
+
dx
(n > 1)
2
2 + b2 )n−1
2(n − 1)b
(x
Z
x
1 2
2
5.
dx
=
ln
x
+
b
2
2
2
Z x +b
x
1
1
6.
n dx = −
2
2
2 (n − 1) (x2 + b2 )n−1
(x + b )
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Proposition. Integrals of the rational functions that cannot be
further
Z decomposed. (Integration constants omitted.)
1
1.
dx = ln |x − c|
Z x−c
1
1
(n > 1)
2.
dx = −
n
(n − 1)(x − c)n−1
Z (x − c)
1
1
x
3.
dx = arctan
2
2
b
b
Z x +b
1
x
+
4.
dx
=
n
2 (x2 + b2 )n−1
(x2 + b2 )
2(n
−
1)b
Z
2n − 3
1
+
dx
(n > 1)
2
2 + b2 )n−1
2(n − 1)b
(x
Z
x
1 2
2
5.
dx
=
ln
x
+
b
2
2
2
Z x +b
x
1
1
6.
(n > 1)
n dx = −
2
2
2 (n − 1) (x2 + b2 )n−1
(x + b )
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Bernd Schröder
Integration by Partial Fraction Decomposition
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
Partial fraction decomposition.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
Partial fraction decomposition.
x+9
(x + 3)(x + 4)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
Partial fraction decomposition.
x+9
A
B
=
+
(x + 3)(x + 4)
x+3 x+4
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
Partial fraction decomposition.
x+9
A
B
=
+
(x + 3)(x + 4)
x+3 x+4
x+9
A(x + 4) + B(x + 3)
=
(x + 3)(x + 4)
(x + 3)(x + 4)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
Partial fraction decomposition.
x+9
A
B
=
+
(x + 3)(x + 4)
x+3 x+4
x+9
A(x + 4) + B(x + 3)
=
(x + 3)(x + 4)
(x + 3)(x + 4)
x + 9 = A(x + 4) + B(x + 3)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
Partial fraction decomposition.
x+9
A
B
=
+
(x + 3)(x + 4)
x+3 x+4
x+9
A(x + 4) + B(x + 3)
=
(x + 3)(x + 4)
(x + 3)(x + 4)
x + 9 = A(x + 4) + B(x + 3)
Heaviside0 s Method :
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
Partial fraction decomposition.
x+9
A
B
=
+
(x + 3)(x + 4)
x+3 x+4
x+9
A(x + 4) + B(x + 3)
=
(x + 3)(x + 4)
(x + 3)(x + 4)
x + 9 = A(x + 4) + B(x + 3)
Heaviside0 s Method :
x = −3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
Partial fraction decomposition.
x+9
A
B
=
+
(x + 3)(x + 4)
x+3 x+4
x+9
A(x + 4) + B(x + 3)
=
(x + 3)(x + 4)
(x + 3)(x + 4)
x + 9 = A(x + 4) + B(x + 3)
Heaviside0 s Method :
x = −3
A=6
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
Partial fraction decomposition.
x+9
A
B
=
+
(x + 3)(x + 4)
x+3 x+4
x+9
A(x + 4) + B(x + 3)
=
(x + 3)(x + 4)
(x + 3)(x + 4)
x + 9 = A(x + 4) + B(x + 3)
Heaviside0 s Method :
x = −3
A=6
x = −4
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
Partial fraction decomposition.
x+9
A
B
=
+
(x + 3)(x + 4)
x+3 x+4
x+9
A(x + 4) + B(x + 3)
=
(x + 3)(x + 4)
(x + 3)(x + 4)
x + 9 = A(x + 4) + B(x + 3)
Heaviside0 s Method :
x = −3
A=6
B = −5
x = −4
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
Partial fraction decomposition.
x+9
A
B
=
+
(x + 3)(x + 4)
x+3 x+4
x+9
A(x + 4) + B(x + 3)
=
(x + 3)(x + 4)
(x + 3)(x + 4)
x + 9 = A(x + 4) + B(x + 3)
Heaviside0 s Method :
x = −3
A=6
B = −5
x = −4
x+9
6
5
=
−
(x + 3)(x + 4)
x+3 x+4
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Bernd Schröder
Integration by Partial Fraction Decomposition
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Z
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
x+9
dx
(x + 3)(x + 4)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Z
x+9
dx =
(x + 3)(x + 4)
Bernd Schröder
Integration by Partial Fraction Decomposition
Z
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
6
5
−
dx
x+3 x+4
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Z
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
x+9
6
5
dx =
−
dx
(x + 3)(x + 4)
x+3 x+4
= 6 ln |x + 3|
Bernd Schröder
Integration by Partial Fraction Decomposition
Z
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Z
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
x+9
6
5
dx =
−
dx
(x + 3)(x + 4)
x+3 x+4
= 6 ln |x + 3| − 5 ln |x + 4|
Bernd Schröder
Integration by Partial Fraction Decomposition
Z
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Z
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
x+9
6
5
dx =
−
dx
(x + 3)(x + 4)
x+3 x+4
= 6 ln |x + 3| − 5 ln |x + 4| + c
Bernd Schröder
Integration by Partial Fraction Decomposition
Z
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Z
Multiple Zeroes
Complex Zeroes
x+9
dx.
(x + 3)(x + 4)
x+9
6
5
dx =
−
dx
(x + 3)(x + 4)
x+3 x+4
= 6 ln |x + 3| − 5 ln |x + 4| + c
Bernd Schröder
Integration by Partial Fraction Decomposition
Z
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
Z
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
Z
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
2x2 + x
− 3
6x4
Bernd Schröder
Integration by Partial Fraction Decomposition
−
7x3
+
8x2
− 10x
+
1
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
3x2
2x2 + x
− 3
6x4
Bernd Schröder
Integration by Partial Fraction Decomposition
−
7x3
+
8x2
− 10x
+
1
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
3x2
2x2 + x
− 3
6x4
−
7x3
+
8x2
6x4
+
3x3
−
9x2
Bernd Schröder
Integration by Partial Fraction Decomposition
− 10x
+
1
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
3x2
2x2 + x
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
Bernd Schröder
Integration by Partial Fraction Decomposition
− 10x
+
1
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
3x2
2x2 + x
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
Bernd Schröder
Integration by Partial Fraction Decomposition
− 10x
+
1
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
3x2
2x2 + x
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
− 10x3
Bernd Schröder
Integration by Partial Fraction Decomposition
+ 17x2
− 10x
+
1
− 10x
+
1
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
2x2 + x
3x2
−
5x
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
− 10x3
Bernd Schröder
Integration by Partial Fraction Decomposition
+ 17x2
− 10x
+
1
− 10x
+
1
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
2x2 + x
3x2
−
5x
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
Bernd Schröder
Integration by Partial Fraction Decomposition
− 10x
+
1
+
1
− 10x3
+ 17x2
− 10x
− 10x3
−
5x2
+ 15x
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
2x2 + x
3x2
−
5x
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
(−)
Bernd Schröder
Integration by Partial Fraction Decomposition
− 10x
+
1
+
1
− 10x3
+ 17x2
− 10x
− 10x3
−
5x2
+ 15x
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
2x2 + x
3x2
−
5x
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
(−)
Bernd Schröder
Integration by Partial Fraction Decomposition
− 10x
+
1
+
1
− 10x3
+ 17x2
− 10x
− 10x3
−
5x2
+ 15x
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
2x2 + x
3x2
−
5x
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
(−)
Bernd Schröder
Integration by Partial Fraction Decomposition
− 10x
+
1
+
1
+
1
− 10x3
+ 17x2
− 10x
− 10x3
−
5x2
+ 15x
22x2
− 25x
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
2x2 + x
3x2
−
5x
+
11
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
(−)
Bernd Schröder
Integration by Partial Fraction Decomposition
− 10x
+
1
+
1
+
1
− 10x3
+ 17x2
− 10x
− 10x3
−
5x2
+ 15x
22x2
− 25x
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
2x2 + x
3x2
−
5x
+
11
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
(−)
Bernd Schröder
Integration by Partial Fraction Decomposition
− 10x
+
1
+
1
1
− 10x3
+ 17x2
− 10x
− 10x3
−
5x2
+ 15x
22x2
− 25x
+
22x2
+ 11x
− 33
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
2x2 + x
3x2
−
5x
+
11
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
(−)
Integration by Partial Fraction Decomposition
+
1
+
1
1
− 10x3
+ 17x2
− 10x
− 10x3
−
5x2
+ 15x
22x2
− 25x
+
22x2
+ 11x
− 33
(−)
Bernd Schröder
− 10x
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
2x2 + x
3x2
−
5x
+
11
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
(−)
Integration by Partial Fraction Decomposition
+
1
+
1
1
− 10x3
+ 17x2
− 10x
− 10x3
−
5x2
+ 15x
22x2
− 25x
+
22x2
+ 11x
− 33
(−)
Bernd Schröder
− 10x
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
2x2 + x
3x2
−
5x
+
11
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
(−)
Integration by Partial Fraction Decomposition
+
1
+
1
1
− 10x3
+ 17x2
− 10x
− 10x3
−
5x2
+ 15x
22x2
− 25x
+
22x2
+ 11x
− 33
− 36x
+ 34
(−)
Bernd Schröder
− 10x
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
2x2 + x
3x2
−
5x
+
11
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
(−)
− 10x
+
1
+
1
1
− 10x3
+ 17x2
− 10x
− 10x3
−
5x2
+ 15x
22x2
− 25x
+
22x2
+ 11x
− 33
− 36x
+ 34
(−)
Thus
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
2x2 + x
3x2
−
5x
+
11
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
(−)
− 10x
+
1
+
1
1
− 10x3
+ 17x2
− 10x
− 10x3
−
5x2
+ 15x
22x2
− 25x
+
22x2
+ 11x
− 33
− 36x
+ 34
(−)
Thus
6x4 − 7x3 + 8x2 − 10x + 1
2x2 + x − 3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
2x2 + x
3x2
−
5x
+
11
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
(−)
− 10x
+
1
+
1
1
− 10x3
+ 17x2
− 10x
− 10x3
−
5x2
+ 15x
22x2
− 25x
+
22x2
+ 11x
− 33
− 36x
+ 34
(−)
Thus
6x4 − 7x3 + 8x2 − 10x + 1
= 3x2 − 5x + 11
2x2 + x − 3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
2x2 + x
3x2
−
5x
+
11
− 3
6x4
−
7x3
+
8x2
(−)
6x4
+
3x3
−
9x2
(−)
− 10x
+
1
+
1
1
− 10x3
+ 17x2
− 10x
− 10x3
−
5x2
+ 15x
22x2
− 25x
+
22x2
+ 11x
− 33
− 36x
+ 34
(−)
Thus
6x4 − 7x3 + 8x2 − 10x + 1
−36x + 34
= 3x2 − 5x + 11 + 2
.
2
2x + x − 3
2x + x − 3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Partial fraction decomposition.
−36x + 34
2x2 + x − 3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Partial fraction decomposition.
−36x + 34
−36x + 34
=
(2x + 3)(x − 1)
2x2 + x − 3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Partial fraction decomposition.
−36x + 34
A
B
−36x + 34
=
=
+
(2x + 3)(x − 1) 2x + 3 x − 1
2x2 + x − 3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Partial fraction decomposition.
−36x + 34
A
B
−36x + 34
=
=
+
(2x + 3)(x − 1) 2x + 3 x − 1
2x2 + x − 3
− 36x + 34 = A(x − 1) + B(2x + 3)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Partial fraction decomposition.
−36x + 34
A
B
−36x + 34
=
=
+
(2x + 3)(x − 1) 2x + 3 x − 1
2x2 + x − 3
− 36x + 34 = A(x − 1) + B(2x + 3)
x=1:
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Partial fraction decomposition.
−36x + 34
A
B
−36x + 34
=
=
+
(2x + 3)(x − 1) 2x + 3 x − 1
2x2 + x − 3
− 36x + 34 = A(x − 1) + B(2x + 3)
x=1:
−2 = B·5
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Partial fraction decomposition.
−36x + 34
A
B
−36x + 34
=
=
+
(2x + 3)(x − 1) 2x + 3 x − 1
2x2 + x − 3
− 36x + 34 = A(x − 1) + B(2x + 3)
2
x = 1 : − 2 = B · 5, B = −
5
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Partial fraction decomposition.
−36x + 34
A
B
−36x + 34
=
=
+
(2x + 3)(x − 1) 2x + 3 x − 1
2x2 + x − 3
− 36x + 34 = A(x − 1) + B(2x + 3)
2
x = 1 : − 2 = B · 5, B = −
5
3
x=− :
2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Partial fraction decomposition.
−36x + 34
A
B
−36x + 34
=
=
+
(2x + 3)(x − 1) 2x + 3 x − 1
2x2 + x − 3
− 36x + 34 = A(x − 1) + B(2x + 3)
2
x = 1 : − 2 = B · 5, B = −
5
3
5
x = − : 88 = A · −
2
2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Partial fraction decomposition.
−36x + 34
A
B
−36x + 34
=
=
+
(2x + 3)(x − 1) 2x + 3 x − 1
2x2 + x − 3
− 36x + 34 = A(x − 1) + B(2x + 3)
2
x = 1 : − 2 = B · 5, B = −
5
3
5
176
x = − : 88 = A · −
, A=−
2
2
5
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Partial fraction decomposition.
−36x + 34
A
B
−36x + 34
=
=
+
(2x + 3)(x − 1) 2x + 3 x − 1
2x2 + x − 3
− 36x + 34 = A(x − 1) + B(2x + 3)
2
x = 1 : − 2 = B · 5, B = −
5
3
5
176
x = − : 88 = A · −
, A=−
2
2
5
−36x + 34
2x2 + x − 3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Partial fraction decomposition.
−36x + 34
A
B
−36x + 34
=
=
+
(2x + 3)(x − 1) 2x + 3 x − 1
2x2 + x − 3
− 36x + 34 = A(x − 1) + B(2x + 3)
2
x = 1 : − 2 = B · 5, B = −
5
3
5
176
x = − : 88 = A · −
, A=−
2
2
5
−36x + 34
176 1
= −
5 2x + 3
2x2 + x − 3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Partial fraction decomposition.
−36x + 34
A
B
−36x + 34
=
=
+
(2x + 3)(x − 1) 2x + 3 x − 1
2x2 + x − 3
− 36x + 34 = A(x − 1) + B(2x + 3)
2
x = 1 : − 2 = B · 5, B = −
5
3
5
176
x = − : 88 = A · −
, A=−
2
2
5
−36x + 34
176 1
2 1
= −
−
5 2x + 3 5 x − 1
2x2 + x − 3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Z
6x4 − 7x3 + 8x2 − 10x + 1
dx
2x2 + x − 3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Z
6x4 − 7x3 + 8x2 − 10x + 1
dx
2x2 + x − 3
Z
=
3x2 − 5x + 11
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Z
6x4 − 7x3 + 8x2 − 10x + 1
dx
2x2 + x − 3
Z
−36x + 34
=
3x2 − 5x + 11 + 2
dx
2x + x − 3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Z
6x4 − 7x3 + 8x2 − 10x + 1
dx
2x2 + x − 3
Z
−36x + 34
=
3x2 − 5x + 11 + 2
dx
2x + x − 3
Z
176 1
2 1
=
3x2 − 5x + 11 −
−
dx
5 2x + 3 5 x − 1
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Z
6x4 − 7x3 + 8x2 − 10x + 1
dx
2x2 + x − 3
Z
−36x + 34
=
3x2 − 5x + 11 + 2
dx
2x + x − 3
Z
176 1
2 1
=
3x2 − 5x + 11 −
−
dx
5 2x + 3 5 x − 1
= x3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Z
6x4 − 7x3 + 8x2 − 10x + 1
dx
2x2 + x − 3
Z
−36x + 34
=
3x2 − 5x + 11 + 2
dx
2x + x − 3
Z
176 1
2 1
=
3x2 − 5x + 11 −
−
dx
5 2x + 3 5 x − 1
5
= x3 − x2
2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Z
6x4 − 7x3 + 8x2 − 10x + 1
dx
2x2 + x − 3
Z
−36x + 34
=
3x2 − 5x + 11 + 2
dx
2x + x − 3
Z
176 1
2 1
=
3x2 − 5x + 11 −
−
dx
5 2x + 3 5 x − 1
5
= x3 − x2 + 11x
2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Z
6x4 − 7x3 + 8x2 − 10x + 1
dx
2x2 + x − 3
Z
−36x + 34
=
3x2 − 5x + 11 + 2
dx
2x + x − 3
Z
176 1
2 1
=
3x2 − 5x + 11 −
−
dx
5 2x + 3 5 x − 1
5
176 1
= x3 − x2 + 11x −
ln |2x + 3|
2
5 2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Z
6x4 − 7x3 + 8x2 − 10x + 1
dx
2x2 + x − 3
Z
−36x + 34
=
3x2 − 5x + 11 + 2
dx
2x + x − 3
Z
176 1
2 1
=
3x2 − 5x + 11 −
−
dx
5 2x + 3 5 x − 1
5
176 1
2
= x3 − x2 + 11x −
ln |2x + 3| − ln |x − 1|
2
5 2
5
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Z
6x4 − 7x3 + 8x2 − 10x + 1
dx
2x2 + x − 3
Z
−36x + 34
=
3x2 − 5x + 11 + 2
dx
2x + x − 3
Z
176 1
2 1
=
3x2 − 5x + 11 −
−
dx
5 2x + 3 5 x − 1
5
176 1
2
= x3 − x2 + 11x −
ln |2x + 3| − ln |x − 1| + c
2
5 2
5
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Z
6x4 − 7x3 + 8x2 − 10x + 1
dx
2x2 + x − 3
Z
−36x + 34
=
3x2 − 5x + 11 + 2
dx
2x + x − 3
Z
176 1
2 1
=
3x2 − 5x + 11 −
−
dx
5 2x + 3 5 x − 1
5
176 1
2
= x3 − x2 + 11x −
ln |2x + 3| − ln |x − 1| + c
2
5 2
5
5
88
2
= x3 − x2 + 11x − ln |2x + 3| − ln |x − 1| + c
2
5
5
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
Complex Zeroes
Example. Compute the integral
6x4 − 7x3 + 8x2 − 10x + 1
dx.
2x2 + x − 3
Z
Z
6x4 − 7x3 + 8x2 − 10x + 1
dx
2x2 + x − 3
Z
−36x + 34
=
3x2 − 5x + 11 + 2
dx
2x + x − 3
Z
176 1
2 1
=
3x2 − 5x + 11 −
−
dx
5 2x + 3 5 x − 1
5
176 1
2
= x3 − x2 + 11x −
ln |2x + 3| − ln |x − 1| + c
2
5 2
5
5
88
2
= x3 − x2 + 11x − ln |2x + 3| − ln |x − 1| + c
2
5
5
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Solve the integral
Bernd Schröder
Integration by Partial Fraction Decomposition
x+3
x2 (x − 1)(x + 2)
Multiple Zeroes
Complex Zeroes
dx.
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
x+3
Z
Solve the integral
f
:=
x2 (x − 1)(x + 2)
Multiple Zeroes
Complex Zeroes
dx.
x+3
x2 (x − 1)(x + 2)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
x+3
Z
Solve the integral
f
:=
x2 (x − 1)(x + 2)
x+3
x2 (x − 1)(x + 2)
Bernd Schröder
Integration by Partial Fraction Decomposition
=
Multiple Zeroes
Complex Zeroes
dx.
A B
D
C
+ 2+
+
x x
x−1 x+2
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
x+3
Z
Solve the integral
f
:=
x+3
=
x2 (x − 1)(x + 2)
x+3
x2 (x − 1)(x + 2)
=
Multiple Zeroes
Complex Zeroes
dx.
A B
D
C
+ 2+
+
x x
x−1 x+2
Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx2 (x + 2) + Dx2 (x − 1)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
x+3
Z
Solve the integral
f
:=
x+3
=
x=0:
x2 (x − 1)(x + 2)
x+3
x2 (x − 1)(x + 2)
=
Multiple Zeroes
Complex Zeroes
dx.
A B
D
C
+ 2+
+
x x
x−1 x+2
Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx2 (x + 2) + Dx2 (x − 1)
3 = B · (−1) · 2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
x+3
Z
Solve the integral
f
:=
x+3
=
x=0:
x2 (x − 1)(x + 2)
x+3
x2 (x − 1)(x + 2)
=
Multiple Zeroes
Complex Zeroes
dx.
A B
D
C
+ 2+
+
x x
x−1 x+2
Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx2 (x + 2) + Dx2 (x − 1)
3
3 = B · (−1) · 2,
B=−
2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
x+3
Z
Solve the integral
f
:=
x+3
=
x=0:
x=1:
x2 (x − 1)(x + 2)
x+3
x2 (x − 1)(x + 2)
=
Multiple Zeroes
Complex Zeroes
dx.
A B
D
C
+ 2+
+
x x
x−1 x+2
Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx2 (x + 2) + Dx2 (x − 1)
3
3 = B · (−1) · 2,
B=−
2
4 = C · 12 · 3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
x+3
Z
Solve the integral
f
:=
x+3
=
x=0:
x=1:
x2 (x − 1)(x + 2)
x+3
x2 (x − 1)(x + 2)
=
Multiple Zeroes
Complex Zeroes
dx.
A B
D
C
+ 2+
+
x x
x−1 x+2
Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx2 (x + 2) + Dx2 (x − 1)
3
3 = B · (−1) · 2,
B=−
2
4
4 = C · 12 · 3,
C=
3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
x+3
Z
Solve the integral
f
:=
x+3
=
x=0:
x=1:
x = −2 :
x2 (x − 1)(x + 2)
x+3
x2 (x − 1)(x + 2)
=
Multiple Zeroes
Complex Zeroes
dx.
A B
D
C
+ 2+
+
x x
x−1 x+2
Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx2 (x + 2) + Dx2 (x − 1)
3
3 = B · (−1) · 2,
B=−
2
4
4 = C · 12 · 3,
C=
3
1 = D · (−2)2 · (−3)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
x+3
Z
Solve the integral
f
:=
x+3
=
x=0:
x=1:
x = −2 :
x2 (x − 1)(x + 2)
x+3
x2 (x − 1)(x + 2)
=
Multiple Zeroes
Complex Zeroes
dx.
A B
D
C
+ 2+
+
x x
x−1 x+2
Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx2 (x + 2) + Dx2 (x − 1)
3
3 = B · (−1) · 2,
B=−
2
4
4 = C · 12 · 3,
C=
3
1
1 = D · (−2)2 · (−3),
D=−
12
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
x+3
Z
Solve the integral
f
x+3
x=0:
x=1:
x = −2 :
x = −1 :
:=
x2 (x − 1)(x + 2)
x+3
x2 (x − 1)(x + 2)
=
Multiple Zeroes
Complex Zeroes
dx.
A B
D
C
+ 2+
+
x x
x−1 x+2
Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx2 (x + 2) + Dx2 (x − 1)
3
3 = B · (−1) · 2,
B=−
2
4
4 = C · 12 · 3,
C=
3
1
1 = D · (−2)2 · (−3),
D=−
12
4 1
2 = A·2+3+ +
3 6
=
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
x+3
Z
Solve the integral
f
x+3
x=0:
x=1:
x = −2 :
x = −1 :
:=
x2 (x − 1)(x + 2)
x+3
x2 (x − 1)(x + 2)
=
Multiple Zeroes
Complex Zeroes
dx.
A B
D
C
+ 2+
+
x x
x−1 x+2
Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx2 (x + 2) + Dx2 (x − 1)
3
3 = B · (−1) · 2,
B=−
2
4
4 = C · 12 · 3,
C=
3
1
1 = D · (−2)2 · (−3),
D=−
12
4 1
5
2 = A·2+3+ + , A = −
3 6
4
=
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
x+3
Z
Solve the integral
f
x+3
x=0:
x=1:
x = −2 :
x = −1 :
f
:=
x2 (x − 1)(x + 2)
x+3
x2 (x − 1)(x + 2)
=
Multiple Zeroes
Complex Zeroes
dx.
A B
D
C
+ 2+
+
x x
x−1 x+2
Ax(x − 1)(x + 2) + B(x − 1)(x + 2) + Cx2 (x + 2) + Dx2 (x − 1)
3
3 = B · (−1) · 2,
B=−
2
4
4 = C · 12 · 3,
C=
3
1
1 = D · (−2)2 · (−3),
D=−
12
4 1
5
2 = A·2+3+ + , A = −
6
3 4
5 1
3
1 4 1
1
1
=
−
+ −
· 2+
+ −
4 x
2
x
3 x−1
12 x + 2
=
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Solve the Integral
Bernd Schröder
Integration by Partial Fraction Decomposition
x+3
x2 (x − 1)(x + 2)
Multiple Zeroes
Complex Zeroes
dx.
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
x+3
Z
Solve the Integral
x+3
Z
x2 (x − 1)(x + 2)
Bernd Schröder
Integration by Partial Fraction Decomposition
x2 (x − 1)(x + 2)
Multiple Zeroes
Complex Zeroes
dx.
dx
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Solve the Integral
x+3
x2 (x − 1)(x + 2)
Multiple Zeroes
Complex Zeroes
dx.
x+3
Z
dx
Z 5 1
3
1 4 1
1
1
=
−
+ −
· 2+
+ −
dx
4 x
2 x
3 x−1
12 x + 2
x2 (x − 1)(x + 2)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Solve the Integral
x+3
x2 (x − 1)(x + 2)
Multiple Zeroes
Complex Zeroes
dx.
x+3
Z
dx
Z 5 1
3
1 4 1
1
1
=
−
+ −
· 2+
+ −
dx
4 x
2 x
3 x−1
12 x + 2
5
= − ln |x|
4
x2 (x − 1)(x + 2)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Solve the Integral
x+3
x2 (x − 1)(x + 2)
Multiple Zeroes
Complex Zeroes
dx.
x+3
Z
dx
Z 5 1
3
1 4 1
1
1
=
−
+ −
· 2+
+ −
dx
4 x
2 x
3 x−1
12 x + 2
5
31
= − ln |x| +
4
2x
x2 (x − 1)(x + 2)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Solve the Integral
x+3
x2 (x − 1)(x + 2)
Multiple Zeroes
Complex Zeroes
dx.
x+3
Z
dx
Z 5 1
3
1 4 1
1
1
=
−
+ −
· 2+
+ −
dx
4 x
2 x
3 x−1
12 x + 2
5
31 4
= − ln |x| +
+ ln |x − 1|
4
2x 3
x2 (x − 1)(x + 2)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Solve the Integral
x+3
x2 (x − 1)(x + 2)
Multiple Zeroes
Complex Zeroes
dx.
x+3
Z
dx
Z 5 1
3
1 4 1
1
1
=
−
+ −
· 2+
+ −
dx
4 x
2 x
3 x−1
12 x + 2
5
31 4
1
= − ln |x| +
+ ln |x − 1| − ln |x + 2|
4
2x 3
12
x2 (x − 1)(x + 2)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Solve the Integral
x+3
x2 (x − 1)(x + 2)
Multiple Zeroes
Complex Zeroes
dx.
x+3
Z
dx
Z 5 1
3
1 4 1
1
1
=
−
+ −
· 2+
+ −
dx
4 x
2 x
3 x−1
12 x + 2
5
31 4
1
= − ln |x| +
+ ln |x − 1| − ln |x + 2| + c
4
2x 3
12
x2 (x − 1)(x + 2)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Solve the Integral
x+3
x2 (x − 1)(x + 2)
Multiple Zeroes
Complex Zeroes
dx.
x+3
Z
dx
Z 5 1
3
1 4 1
1
1
=
−
+ −
· 2+
+ −
dx
4 x
2 x
3 x−1
12 x + 2
5
31 4
1
= − ln |x| +
+ ln |x − 1| − ln |x + 2| + c
4
2x 3
12
x2 (x − 1)(x + 2)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Bernd Schröder
Integration by Partial Fraction Decomposition
Multiple Zeroes
1
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Multiple Zeroes
1
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Multiple Zeroes
1
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
(x2 + 1) (x + 3)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Multiple Zeroes
1
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
Ax + B
=
(x2 + 1) (x + 3)
x2 + 1
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
1
Z
Example. Compute the integral
Multiple Zeroes
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
Ax + B
C
=
+
(x2 + 1) (x + 3)
x2 + 1 x + 3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
1
Z
Example. Compute the integral
Multiple Zeroes
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
Ax + B
C
=
+
(x2 + 1) (x + 3)
x2 + 1 x + 3
2
1 = (Ax + B)(x + 3) + C x + 1
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
1
Z
Example. Compute the integral
Multiple Zeroes
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
Ax + B
C
=
+
(x2 + 1) (x + 3)
x2 + 1 x + 3
2
1 = (Ax + B)(x + 3) + C x + 1
x = −3 :
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
1
Z
Example. Compute the integral
Multiple Zeroes
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
Ax + B
C
=
+
(x2 + 1) (x + 3)
x2 + 1 x + 3
2
1 = (Ax + B)(x + 3) + C x + 1
x = −3 :
Bernd Schröder
Integration by Partial Fraction Decomposition
1 = 10C
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
1
Z
Example. Compute the integral
Multiple Zeroes
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
Ax + B
C
=
+
(x2 + 1) (x + 3)
x2 + 1 x + 3
2
1 = (Ax + B)(x + 3) + C x + 1
x = −3 :
Bernd Schröder
Integration by Partial Fraction Decomposition
1 = 10C,
C=
1
10
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
1
Z
Example. Compute the integral
Multiple Zeroes
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
Ax + B
C
=
+
(x2 + 1) (x + 3)
x2 + 1 x + 3
2
1 = (Ax + B)(x + 3) + C x + 1
x = −3 :
1 = 10C,
C=
1
10
x=0:
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
1
Z
Example. Compute the integral
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
Ax + B
C
=
+
(x2 + 1) (x + 3)
x2 + 1 x + 3
2
1 = (Ax + B)(x + 3) + C x + 1
x = −3 :
1 = 10C,
x=0:
1 = 3B +
Bernd Schröder
Integration by Partial Fraction Decomposition
C=
1
10
1
10
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
1
Z
Example. Compute the integral
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
Ax + B
C
=
+
(x2 + 1) (x + 3)
x2 + 1 x + 3
2
1 = (Ax + B)(x + 3) + C x + 1
x = −3 :
1 = 10C,
x=0:
1 = 3B +
Bernd Schröder
Integration by Partial Fraction Decomposition
C=
1
,
10
1
10
B=
3
10
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Multiple Zeroes
1
Z
Example. Compute the integral
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
Ax + B
C
=
+
(x2 + 1) (x + 3)
x2 + 1 x + 3
2
1 = (Ax + B)(x + 3) + C x + 1
x = −3 :
1 = 10C,
x=0:
1 = 3B +
C=
1
,
10
1
10
B=
3
10
x=1:
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
1
Z
Example. Compute the integral
Multiple Zeroes
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
Ax + B
C
=
+
(x2 + 1) (x + 3)
x2 + 1 x + 3
2
1 = (Ax + B)(x + 3) + C x + 1
x = −3 :
x=0:
x=1:
Bernd Schröder
Integration by Partial Fraction Decomposition
1 = 10C,
C=
1
10
1
3
1 = 3B + ,
B=
10 10
3
1
1 = A+
4+ ·2
10
10
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
1
Z
Example. Compute the integral
Multiple Zeroes
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
Ax + B
C
=
+
(x2 + 1) (x + 3)
x2 + 1 x + 3
2
1 = (Ax + B)(x + 3) + C x + 1
x = −3 :
x=0:
x=1:
Bernd Schröder
Integration by Partial Fraction Decomposition
1 = 10C,
C=
1
10
1
3
1 = 3B + ,
B=
10 10
3
1
1 = A+
4 + · 2,
10
10
A=−
1
10
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
1
Z
Example. Compute the integral
Multiple Zeroes
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
Ax + B
C
=
+
(x2 + 1) (x + 3)
x2 + 1 x + 3
2
1 = (Ax + B)(x + 3) + C x + 1
x = −3 :
1 = 10C,
x=1:
1
1
10
1
3
1 = 3B + ,
B=
10 10
3
1
1 = A+
4 + · 2,
10
10
x=0:
(x2 + 1) (x + 3)
C=
A=−
1
10
=
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
1
Z
Example. Compute the integral
Multiple Zeroes
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
Ax + B
C
=
+
(x2 + 1) (x + 3)
x2 + 1 x + 3
2
1 = (Ax + B)(x + 3) + C x + 1
x = −3 :
1 = 10C,
C=
1
10
1
3
1 = 3B + ,
B=
10 10
3
1
x=1:
1 = A+
4 + · 2,
10
10
1
1 −x + 3
=
2
10 x2 + 1
(x + 1) (x + 3)
x=0:
Bernd Schröder
Integration by Partial Fraction Decomposition
A=−
1
10
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
1
Z
Example. Compute the integral
Multiple Zeroes
(x2 + 1) (x + 3)
Complex Zeroes
dx.
Partial fraction decomposition.
1
Ax + B
C
=
+
(x2 + 1) (x + 3)
x2 + 1 x + 3
2
1 = (Ax + B)(x + 3) + C x + 1
x = −3 :
1 = 10C,
C=
1
10
1
3
1 = 3B + ,
B=
10 10
3
1
x=1:
1 = A+
4 + · 2,
10
10
1
1 −x + 3
1 1
=
+
2
2
10 x + 1 10 x + 3
(x + 1) (x + 3)
x=0:
Bernd Schröder
Integration by Partial Fraction Decomposition
A=−
1
10
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
Z
1
(x2 + 1) (x + 3)
Bernd Schröder
Integration by Partial Fraction Decomposition
Multiple Zeroes
1
(x2 + 1) (x + 3)
Complex Zeroes
dx.
dx
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
1
Z
Example. Compute the integral
1
Z
(x2 + 1) (x + 3)
Z
=
Multiple Zeroes
(x2 + 1) (x + 3)
Complex Zeroes
dx.
dx
1 −x + 3
1 1
+
dx
2
10 x + 1 10 x + 3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
1
Z
(x2 + 1) (x + 3)
Multiple Zeroes
1
(x2 + 1) (x + 3)
Complex Zeroes
dx.
dx
1 −x + 3
1 1
+
dx
2
10 x + 1 10 x + 3
Z
1 x
3 1
1 1
=
+
+
dx
−
2
2
10 x + 1 10 x + 1 10 x + 3
Z
=
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
1
Z
(x2 + 1) (x + 3)
Multiple Zeroes
1
(x2 + 1) (x + 3)
Complex Zeroes
dx.
dx
1 −x + 3
1 1
+
dx
2
10 x + 1 10 x + 3
Z
1 x
3 1
1 1
=
+
+
dx
−
2
2
10 x + 1 10 x + 1 10 x + 3
1 1 2
= −
ln x + 1
10 2
Z
=
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
1
Z
(x2 + 1) (x + 3)
Multiple Zeroes
1
(x2 + 1) (x + 3)
Complex Zeroes
dx.
dx
1 −x + 3
1 1
+
dx
2
10 x + 1 10 x + 3
Z
1 x
3 1
1 1
=
+
+
dx
−
2
2
10 x + 1 10 x + 1 10 x + 3
1 1 2
3
= −
ln x + 1 + arctan(x)
10 2
10
Z
=
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
1
Z
(x2 + 1) (x + 3)
Multiple Zeroes
1
(x2 + 1) (x + 3)
Complex Zeroes
dx.
dx
1 −x + 3
1 1
+
dx
2
10 x + 1 10 x + 3
Z
1 x
3 1
1 1
=
+
+
dx
−
2
2
10 x + 1 10 x + 1 10 x + 3
1 1 2
3
1
= −
ln x + 1 + arctan(x) + ln |x + 3|
10 2
10
10
Z
=
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
1
Z
(x2 + 1) (x + 3)
Multiple Zeroes
1
(x2 + 1) (x + 3)
Complex Zeroes
dx.
dx
1 −x + 3
1 1
+
dx
2
10 x + 1 10 x + 3
Z
1 x
3 1
1 1
=
+
+
dx
−
2
2
10 x + 1 10 x + 1 10 x + 3
1 1 2
3
1
= −
ln x + 1 + arctan(x) + ln |x + 3| + c
10 2
10
10
Z
=
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
1
Z
(x2 + 1) (x + 3)
Multiple Zeroes
1
(x2 + 1) (x + 3)
Complex Zeroes
dx.
dx
1 −x + 3
1 1
+
dx
2
10 x + 1 10 x + 3
Z
1 x
3 1
1 1
=
+
+
dx
−
2
2
10 x + 1 10 x + 1 10 x + 3
1 1 2
3
1
= −
ln x + 1 + arctan(x) + ln |x + 3| + c
10 2
10
10
3
1 2
1
= − ln x + 1 + arctan(x) + ln |x + 3| + c
20
10
10
Z
=
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Compute the integral
1
Z
(x2 + 1) (x + 3)
Multiple Zeroes
1
(x2 + 1) (x + 3)
Complex Zeroes
dx.
dx
1 −x + 3
1 1
+
dx
2
10 x + 1 10 x + 3
Z
1 x
3 1
1 1
=
+
+
dx
−
2
2
10 x + 1 10 x + 1 10 x + 3
1 1 2
3
1
= −
ln x + 1 + arctan(x) + ln |x + 3| + c
10 2
10
10
3
1 2
1
= − ln x + 1 + arctan(x) + ln |x + 3| + c
20
10
10
Z
=
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Bernd Schröder
Integration by Partial Fraction Decomposition
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
x3 + 2x2 + 4x + 10
Y := 2
=
(x + 2x + 2) (x2 + 4)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
x3 + 2x2 + 4x + 10
Ax + B
Cx + D
Y := 2
= 2
+ 2
2
(x + 2x + 2) (x + 4)
x + 2x + 2 x + 4
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
x3 + 2x2 + 4x + 10
Ax + B
Cx + D
Y := 2
= 2
+ 2
2
(x + 2x + 2) (x + 4)
x + 2x + 2 x + 4
x3 + 2x2 + 4x + 10 =
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
x3 + 2x2 + 4x + 10
Ax + B
Cx + D
Y := 2
= 2
+ 2
2
(x + 2x + 2) (x + 4)
x + 2x + 2 x + 4
3
2
2
2
x + 2x + 4x + 10 = (Ax+B) x +4 + (Cx+D) x +2x+2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
x3 + 2x2 + 4x + 10
Ax + B
Cx + D
Y := 2
= 2
+ 2
2
(x + 2x + 2) (x + 4)
x + 2x + 2 x + 4
3
2
2
2
x + 2x + 4x + 10 = (Ax+B) x +4 + (Cx+D) x +2x+2
x3 + 2x2 + 4x + 10 = (A + C)x3
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
x3 + 2x2 + 4x + 10
Ax + B
Cx + D
Y := 2
= 2
+ 2
2
(x + 2x + 2) (x + 4)
x + 2x + 2 x + 4
3
2
2
2
x + 2x + 4x + 10 = (Ax+B) x +4 + (Cx+D) x +2x+2
x3 + 2x2 + 4x + 10 = (A + C)x3 + (B + 2C + D)x2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
x3 + 2x2 + 4x + 10
Ax + B
Cx + D
Y := 2
= 2
+ 2
2
(x + 2x + 2) (x + 4)
x + 2x + 2 x + 4
3
2
2
2
x + 2x + 4x + 10 = (Ax+B) x +4 + (Cx+D) x +2x+2
x3 + 2x2 + 4x + 10 = (A + C)x3 + (B + 2C + D)x2
+(4A + 2C + 2D)x
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
x3 + 2x2 + 4x + 10
Ax + B
Cx + D
Y := 2
= 2
+ 2
2
(x + 2x + 2) (x + 4)
x + 2x + 2 x + 4
3
2
2
2
x + 2x + 4x + 10 = (Ax+B) x +4 + (Cx+D) x +2x+2
x3 + 2x2 + 4x + 10 = (A + C)x3 + (B + 2C + D)x2
+(4A + 2C + 2D)x + (4B + 2D)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
x3 + 2x2 + 4x + 10
Ax + B
Cx + D
Y := 2
= 2
+ 2
2
(x + 2x + 2) (x + 4)
x + 2x + 2 x + 4
3
2
2
2
x + 2x + 4x + 10 = (Ax+B) x +4 + (Cx+D) x +2x+2
x3 + 2x2 + 4x + 10 = (A + C)x3 + (B + 2C + D)x2
+(4A + 2C + 2D)x + (4B + 2D)
A+C = 1
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
x3 + 2x2 + 4x + 10
Ax + B
Cx + D
Y := 2
= 2
+ 2
2
(x + 2x + 2) (x + 4)
x + 2x + 2 x + 4
3
2
2
2
x + 2x + 4x + 10 = (Ax+B) x +4 + (Cx+D) x +2x+2
x3 + 2x2 + 4x + 10 = (A + C)x3 + (B + 2C + D)x2
+(4A + 2C + 2D)x + (4B + 2D)
A+C = 1
B + 2C + D = 2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
x3 + 2x2 + 4x + 10
Ax + B
Cx + D
Y := 2
= 2
+ 2
2
(x + 2x + 2) (x + 4)
x + 2x + 2 x + 4
3
2
2
2
x + 2x + 4x + 10 = (Ax+B) x +4 + (Cx+D) x +2x+2
x3 + 2x2 + 4x + 10 = (A + C)x3 + (B + 2C + D)x2
+(4A + 2C + 2D)x + (4B + 2D)
A+C = 1
B + 2C + D = 2
4A + 2C + 2D = 4
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
x3 + 2x2 + 4x + 10
Ax + B
Cx + D
Y := 2
= 2
+ 2
2
(x + 2x + 2) (x + 4)
x + 2x + 2 x + 4
3
2
2
2
x + 2x + 4x + 10 = (Ax+B) x +4 + (Cx+D) x +2x+2
x3 + 2x2 + 4x + 10 = (A + C)x3 + (B + 2C + D)x2
+(4A + 2C + 2D)x + (4B + 2D)
A+C = 1
B + 2C + D = 2
4A + 2C + 2D = 4
4B + 2D = 10
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Bernd Schröder
Integration by Partial Fraction Decomposition
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
A
+
B
4A
+
4B
C
+ 2C
2C
+
+ D
+ 2D
2D
Bernd Schröder
Integration by Partial Fraction Decomposition
=
=
=
=
1
2
4
10
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
A
+
B
4A
+
4B
C
+ 2C
2C
+
+ D
+ 2D
2D
Bernd Schröder
Integration by Partial Fraction Decomposition
=
=
=
=
1
2
4
10
A
+
B
4B
C
+ 2C
− 2C
+
+ D
+ 2D
2D
=
=
=
=
1
2
0
10
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
A
+
B
4A
+
4B
A
+
B
C
+ 2C
2C
+
C
+ 2C
− 2C
− 8C
=
=
=
=
+ D
+ 2D
2D
+ D
+ 2D
− 2D
Bernd Schröder
Integration by Partial Fraction Decomposition
=
=
=
=
1
2
4
10
A
+
B
4B
C
+ 2C
− 2C
+
+ D
+ 2D
2D
=
=
=
=
1
2
0
10
1
2
0
2
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
A
+
B
4A
+
4B
A
+
B
C
+ 2C
2C
+
C
+ 2C
− 2C
− 8C
=
=
=
=
+ D
+ 2D
2D
+ D
+ 2D
− 2D
Bernd Schröder
Integration by Partial Fraction Decomposition
=
=
=
=
1
2
0
2
1
2
4
10
A
+
B
4B
A
+
B
C
+ 2C
− 2C
+
C
+ 2C
− 2C
+ D
+ 2D
2D
+ D
+ 2D
− 10D
=
=
=
=
1
2
0
10
=
=
=
=
1
2
0
2
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
A
+
C
+ 2C
2C
+
B
4A
+
4B
A
+
B
C
+ 2C
− 2C
− 8C
=
=
=
=
+ D
+ 2D
2D
+ D
+ 2D
− 2D
=
=
=
=
1
2
0
2
1
2
4
10
A
+
B
4B
A
+
B
C
+ 2C
− 2C
+
C
+ 2C
− 2C
+ D
+ 2D
2D
+ D
+ 2D
− 10D
=
=
=
=
1
2
0
10
=
=
=
=
1
2
0
2
1
D=− ,
5
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
A
+
B
4A
+
4B
A
+
B
C
+ 2C
2C
+
C
+ 2C
− 2C
− 8C
=
=
=
=
+ D
+ 2D
2D
+ D
+ 2D
− 2D
=
=
=
=
1
2
0
2
1
2
4
10
A
+
B
4B
A
+
B
C
+ 2C
− 2C
+
C
+ 2C
− 2C
+ D
+ 2D
2D
+ D
+ 2D
− 10D
=
=
=
=
1
2
0
10
=
=
=
=
1
2
0
2
1
1
D=− ,C=− ,
5
5
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
A
+
B
4A
+
4B
A
+
B
C
+ 2C
2C
+
C
+ 2C
− 2C
− 8C
=
=
=
=
+ D
+ 2D
2D
+ D
+ 2D
− 2D
=
=
=
=
1
2
0
2
1
2
4
10
A
+
B
4B
A
+
B
C
+ 2C
− 2C
+
C
+ 2C
− 2C
+ D
+ 2D
2D
+ D
+ 2D
− 10D
=
=
=
=
1
2
0
10
=
=
=
=
1
2
0
2
1
1
13
D=− ,C=− ,B= ,
5
5
5
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
A
+
B
4A
+
4B
A
+
B
C
+ 2C
2C
+
C
+ 2C
− 2C
− 8C
=
=
=
=
+ D
+ 2D
2D
+ D
+ 2D
− 2D
=
=
=
=
1
2
4
10
1
2
0
2
A
+
B
4B
A
+
B
C
+ 2C
− 2C
+
C
+ 2C
− 2C
+ D
+ 2D
2D
+ D
+ 2D
− 10D
=
=
=
=
1
2
0
10
=
=
=
=
1
2
0
2
1
1
13
6
D=− ,C=− ,B= ,A= ,
5
5
5
5
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Partial fraction decomposition.
A
+
B
4A
+
4B
A
+
B
C
+ 2C
2C
+
C
+ 2C
− 2C
− 8C
=
=
=
=
+ D
+ 2D
2D
+ D
+ 2D
− 2D
=
=
=
=
1
2
4
10
1
2
0
2
1
1
13
6
D=− ,C=− ,B= ,A= ,
5
5
5
5
Bernd Schröder
Integration by Partial Fraction Decomposition
+
A
B
4B
A
+
B
Y=
C
+ 2C
− 2C
+
C
+ 2C
− 2C
6
13
5x+ 5
x2 + 2x + 2
+
+ D
+ 2D
2D
+ D
+ 2D
− 10D
=
=
=
=
1
2
0
10
=
=
=
=
1
2
0
2
− 15 x − 15
x2 + 4
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Bernd Schröder
Integration by Partial Fraction Decomposition
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Z
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Z
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Z
6
13
− 51 x − 15
5x + 5
=
+ 2
dx
x2 + 2x + 2
x +4
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Z
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Z
6
13
− 51 x − 15
5x + 5
=
+ 2
dx
x2 + 2x + 2
x +4
Z
6
x+1
7
1
1 x
1 1
=
+
− 2
− 2
dx
2
2
5 (x + 1) + 1 5 (x + 1) + 1 5 x + 4 5 x + 4
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Z
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Z
6
13
− 51 x − 15
5x + 5
=
+ 2
dx
x2 + 2x + 2
x +4
Z
6
x+1
7
1
1 x
1 1
=
+
− 2
− 2
dx
2
2
5 (x + 1) + 1 5 (x + 1) + 1 5 x + 4 5 x + 4
61 =
ln (x + 1)2 + 1
52
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Z
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Z
6
13
− 51 x − 15
5x + 5
=
+ 2
dx
x2 + 2x + 2
x +4
Z
6
x+1
7
1
1 x
1 1
=
+
− 2
− 2
dx
2
2
5 (x + 1) + 1 5 (x + 1) + 1 5 x + 4 5 x + 4
7
61 =
ln (x + 1)2 + 1 + arctan (x + 1)
5
52
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Z
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Z
6
13
− 51 x − 15
5x + 5
=
+ 2
dx
x2 + 2x + 2
x +4
Z
6
x+1
7
1
1 x
1 1
=
+
− 2
− 2
dx
2
2
5 (x + 1) + 1 5 (x + 1) + 1 5 x + 4 5 x + 4
7
11 2
61 ln x + 4
=
ln (x + 1)2 + 1 + arctan (x + 1) −
5
52
52
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Z
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Z
6
13
− 51 x − 15
5x + 5
=
+ 2
dx
x2 + 2x + 2
x +4
Z
6
x+1
7
1
1 x
1 1
=
+
− 2
− 2
dx
2
2
5 (x + 1) + 1 5 (x + 1) + 1 5 x + 4 5 x + 4
7
11 2
61 ln x + 4
=
ln (x + 1)2 + 1 + arctan (x + 1) −
5
52
52
x
11
−
arctan
52
2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Z
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Z
6
13
− 51 x − 15
5x + 5
=
+ 2
dx
x2 + 2x + 2
x +4
Z
6
x+1
7
1
1 x
1 1
=
+
− 2
− 2
dx
2
2
5 (x + 1) + 1 5 (x + 1) + 1 5 x + 4 5 x + 4
7
11 2
61 ln x + 4
=
ln (x + 1)2 + 1 + arctan (x + 1) −
5
52
52
x
11
−
arctan
+c
52
2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Bernd Schröder
Integration by Partial Fraction Decomposition
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Z
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Z
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
7
61 11 2
=
ln (x + 1)2 + 1 + arctan (x + 1) −
ln x + 4
52
5
52
x
11
−
arctan
+c
52
2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Z
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
7
61 11 2
=
ln (x + 1)2 + 1 + arctan (x + 1) −
ln x + 4
52
5
52
x
11
−
arctan
+c
52
2
7
3 1 =
ln (x + 1)2 + 1 + arctan (x + 1) − ln x2 + 4
5
5
10
x
1
+c
− arctan
10
2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science
Background
Single Zeroes
Higher Degree Numerator
Z
Example. Solve the integral
Z
Multiple Zeroes
Complex Zeroes
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
x3 + 2x2 + 4x + 10
dx
(x2 + 2x + 2) (x2 + 4)
7
61 11 2
=
ln (x + 1)2 + 1 + arctan (x + 1) −
ln x + 4
52
5
52
x
11
−
arctan
+c
52
2
7
3 1 =
ln (x + 1)2 + 1 + arctan (x + 1) − ln x2 + 4
5
5
10
x
1
+c
− arctan
10
2
Bernd Schröder
Integration by Partial Fraction Decomposition
Louisiana Tech University, College of Engineering and Science