Integration of Rational Expressions by Partial Fractions
INTRODUCTION:
We start with a few definitions. A rational expression is formed when a polynomial is divided
by another polynomial. In a proper rational expression the degree of the numerator is less than
the degree of the denominator. In an improper rational expression the degree of the numerator
is greater than or equal to the degree of the denominator.
This set of notes is given in three parts. Part A is an explanation of how to decompose a proper
rational expression into a sum of simpler fractions. Part B explains Integration by Partial
Fractions of proper rational expressions. Part C explains Integration by Partial Fractions of
improper rational expressions. Each part includes detailed examples and a set of exercises.
PART A: Partial Fraction Decomposition
In mathematics we often combine two or more rational expressions into one.
E.g.
4 (x − 2)
3 ( x + 1)
7x − 5
4
3
=
+
=
+
(x + 1)(x − 2) (x − 2)(x + 1) (x − 2)(x + 1)
x +1 x − 2
Occasionally, however, the reverse procedure is necessary. The problem is to take a fraction
whose denominator is a product of factors, and split it into a sum of simpler fractions. There is
more than one way to do this. For the types of expressions we are dealing with the method
illustrated here is probably the easiest to apply and understand (no system of equations to solve).
CASE 1 The denominator is a product of distinct linear factors.
For each distinct factor ax + b the sum of partial fractions includes a term of the form
Example 1
Rewrite
A
.
ax + b
x+5
as a sum of simpler fractions.
(x − 4)(x −1)
First write the fraction as
x+5
A
B
=
+
.
(x − 4)(x −1) x − 4 x −1
Multiply both sides by the common denominator to get x + 5 = A ( x − 1) + B( x − 4 ) . In this
linear equation we can substitute any x-values to solve for A and B. However, the process is
simpler if we choose those values which make a factor zero ( i.e. x − 1 = 0 or x − 4 = 0 ) .
Substitute x =1 in the linear equation.
1 + 5 = A (1 −1) + B (1 − 4 ) → 6 = −3B → B = −2
Substitute x = 4 in the linear equation.
4 + 5 = A (4 −1) + B(4 − 4 ) → 9 = 3 A → A = 3
Therefore, the partial fraction decomposition is
x+5
3
2
=
−
.
(x − 4)(x −1) x − 4 x −1
Example 2
Rewrite
2x − 2
as a sum of simpler fractions.
(x + 5)(x + 2)(x − 3)
First write the fraction as
2x − 2
A
B
C
=
+
+
.
(x + 5)(x + 2)(x − 3) x + 5 x + 2 x − 3
Multiply both sides by the common denominator to get the linear equation
2 x − 2 = A ( x + 2 )( x − 3) + B ( x + 5)( x − 3) + C ( x + 5)( x + 2 ) .
− 4 − 2 = 0 + B (3)(− 5) + 0 → − 6 = −15B → B = 52
Substitute x = −2 .
Substitute x = 3 .
1
6 − 2 = 0 + 0 + C (8)(5) → 4 = 40 C → C = 10
− 12 = A (− 3)(− 8) + 0 + 0 → − 12 = 24 A → A = − 12
Substitute x = −5 .
The partial fraction decomposition is
2x − 2
1
1
1
= − 1⋅
+ 2⋅
+ 1⋅
.
2 x+5
5 x+2
10 x − 3
(x + 5)(x + 2)(x − 3)
CASE 2 The denominator is a product of linear factors, some of which are repeated
For each repeated linear factor ( ax + b ) , the sum of partial fractions includes n terms of the
n
form
A3
An
A1
A2
+
+
+ ... +
.
2
3
ax + b (ax + b )
(ax + b )
(ax + b )n
Example 3
Rewrite
4x
( x − 2)
First write the fraction as
Multiply both sides by
Substitute x = 2 .
as a sum of simpler fractions.
2
4x
( x − 2)
( x − 2)
2
2
=
A
B
.
+
x − 2 ( x − 2 )2
to get
4 x = A (x − 2) + B .
8 = 0 + B → B =8
There is no other factor to make zero, so we choose an easy x-value to work with.
Substitute x = 0 ( and B = 8 ).
0 = A (− 2 ) + 8 → A = 4
The partial fraction decomposition is
Example 4
4x
(x − 2)
2
=
4
8
.
+
x − 2 ( x − 2 )2
x 2 − 2x − 5
as a sum of simpler fractions.
Rewrite
x 3 − 5x 2
(continued next page)
Page 2 of 12
First write the fraction as
x 2 − 2x − 5
x 2 − 2x − 5
A
B
C
=
=
+ 2 +
.
3
2
2
x
x −5
x − 5x
x ( x − 5)
x
Multiply by x 2 ( x − 5) to get
x 2 − 2 x − 5 = A x ( x − 5) + B ( x − 5) + C x 2 .
Substitute x = 0 .
− 5 = 0 + B (− 5) + 0 → B = 1
Substitute x = 5 .
10 = 0 + 0 + C (25) → C = 52
Substitute x = 1 .
− 6 = A (− 4) + (− 4 ) + 52 (1) → A = 53
The partial fraction decomposition is
x 2 − 2x − 5
1
1
1
= 3⋅ + 2 + 2⋅
.
3
2
5
5
x
x −5
x − 5x
x
CASE 3 The denominator has one or more distinct, irreducible quadratic factors.
For each distinct factor ax 2 + bx + c the sum of partial fractions includes a term
Example 5
Rewrite
x 2 + 4 x + 12
as a sum of simpler fractions.
(x − 2) (x 2 + 4)
First write the fraction as
x 2 + 4 x + 12
A
Bx + C
=
+ 2
.
2
(x − 2) (x + 4) x − 2 x + 4
(x − 2) (x 2 + 4) to get x 2 + 4 x + 12 =
x = 2 . 24 = A (8) + 0 → A = 3
x = 0 . 12 = 3 (4 ) + (C )(− 2 ) → C = 0
x =1 . 17 = 3 (5) + (B )(− 1) → B = −2
Multiply by
Substitute
Substitute
Substitute
Ax + B
.
ax 2 + bx + c
A (x 2 + 4) + (Bx + C )( x − 2) .
x 2 + 4 x + 12
3
2x
=
− 2
The partial fraction decomposition is
.
2
(x − 2) (x + 4) x − 2 x + 4
Example 6
Rewrite
x2 + x −3
as a sum of simpler fractions.
(x + 1) (x 2 − 2 x + 3)
x2 + x −3
A
Bx + C
=
+ 2
.
First write the fraction as
2
(x + 1) (x − 2 x + 3) x + 1 x − 2 x + 3
Multiply by
(x + 1) (x 2 − 2 x + 3)
Substitute x = −1 .
to get
x 2 + x − 3 = A (x 2 − 2 x + 3) + (Bx + C )( x + 1) .
− 3 = A (6 ) + 0 → A = − 12
(continued next page)
Page 3 of 12
− 3 = − 12 (3) + (C )(1) → C = − 32
Substitute x = 0 ( and A = − 12 ).
Substitute x =1 ( and A = − 12 and C = − 32 ).
(
)
− 1 = − 12 (2 ) + B − 32 (2 ) → B = 32
x2 + x −3
1
x −1
= −1⋅
+ 3⋅ 2
.
2
2 x +1 2 x − 2 x + 3
(x + 1) (x − 2 x + 3)
The partial fraction decomposition is
EXERCISES Rewrite each of the following as a sum of simpler fractions.
1.
4
2.
(x + 1)(x − 5)
5x +1
4.
(x + 3)(x + 2)(x − 4)
7.
2x − 3
x − 5x + 6
2
4x 2 − x + 3
5.
(x + 5)(x −1)(x − 2)
x+2
2
x + 8 x + 16
8.
x2 − 2
10.
(x + 1) (x 2 + 3)
5 − 4x
3
x + 10 x 2 + 25 x
3x 2 + 9 x − 4
(x −1) (x 2 + 4 x −1)
11.
3.
5x
2 x + 11x + 12
6.
2x 2 − 5
x 3 − 2 x 2 − 3x
9.
12.
2
5x 2 − 9 x
(x − 4)(x −1)2
6
(x − 5) (x 2 − 2 x + 3)
SOLUTIONS
1.
4
(x + 1)(x − 5)
=
x = 5 → B = 23
A
B
+
x +1 x − 5
and
→
4 = A ( x − 5) + B ( x + 1) for all x
x = −1 → A = − 23
4
1
1
= 2⋅
− 2⋅
3
3
(x + 1)(x − 5)
x −5
x +1
2.
2x − 3
2x − 3
A
B
=
=
+
(x − 2)(x − 3) x − 2 x − 3
x − 5x + 6
2
x =3 → B = 3
and
→
2 x − 3 = A ( x − 3) + B ( x − 2) for all x
x = 2 → A = −1
2x − 3
3
1
=
−
x −3 x −2
x − 5x + 6
2
Page 4 of 12
3.
5x
5x
A
B
=
=
+
(2 x + 3)(x + 4) 2 x + 3 x + 4
2 x + 11x + 12
→
2
x=− 4 → B = 4
and
5 x = A ( x + 4 ) + B (2 x + 3)
x = − 32 → A = −3
5x
4
3
=
−
x + 4 2x + 3
2 x + 11x + 12
2
4.
5x +1
A
B
C
=
+
+
(x + 3)(x + 2)(x − 4) x + 3 x + 2 x − 4
→ 5 x + 1 = A ( x + 2 )( x − 4 ) + B ( x + 3)( x − 4 ) + C ( x + 3)( x + 2 )
x = −2 → B = 32
and
x = 4 → C = 12
x = −3 → A = −2
and
5x +1
2
B
C
= −
+ 3⋅
+ 1⋅
2 x+2
2 x−4
(x + 3)(x + 2)(x − 4)
x+3
4x 2 − x + 3
A
B
C
5.
=
+
+
(x + 5)(x −1)(x − 2) x + 5 x −1 x − 2
→ 4 x 2 − x + 3 = A ( x − 1)( x − 2 ) + B ( x + 5)( x − 2) + C ( x + 5)( x − 1)
x = 1 → B = −1
and
x = 2 → C = 17
7
x = −5 → A = 18
7
and
4x 2 − x + 3
1
1
1
= 18 ⋅
−
+ 17 ⋅
(x + 5)(x −1)(x − 2) 7 x + 5 x −1 7 x − 2
6.
2x 2 − 5
2x 2 − 5
A
B
C
=
=
+
+
3
2
x ( x + 1)( x − 3)
x
x +1 x − 3
x − 2 x − 3x
→ 2 x 2 − 5 = A (x + 1)( x − 3) + B x ( x − 3) + C x ( x + 1)
x = −1 → B = − 34
and
13
x = 3 → C = 12
and
x = 0 → A = 53
2x 2 − 5
1
1
1
= 5⋅ − 3⋅
+ 13 ⋅
3
2
3 x
4 x +1
12 x − 3
x − 2 x − 3x
7.
x+2
x+2
A
B
=
=
+
2
x + 4 ( x + 4 )2
x + 8 x + 16
(x + 4)
2
x = −4 → B = −2
and
→
x + 2 = A (x + 4) + B
x = 0, B = −2 → A = 1
2x 2 − 5
1
1
1
= 9⋅
+ 1⋅
+ 13 ⋅
(x + 4)(x + 1)(x − 3) 7 x + 4 4 x + 1 28 x − 3
Page 5 of 12
8.
5 − 4x
5 − 4x
A
B
C
=
=
+
+
2
2
x
x + 5 ( x + 5 )2
x + 10 x + 25 x
x ( x + 5)
3
→ 5 − 4 x = A ( x + 5) + B x ( x + 5) + C x
2
x = −5 → C = −5
and
x = 0 → A = 15
and
x =1, A = 15 , C = −5 → B = − 15
5 − 4x
1
5
1
= 1⋅
− 1⋅
−
2
5 x+5
5 x −3
x + 10 x + 25
( x + 5 )2
3
9.
5x 2 − 9 x
(x − 4)(x −1)
2
A
B
C
+
+
x − 4 x −1 (x −1)2
=
→ 5 x 2 − 9 x = A ( x − 1) + B ( x − 4)( x − 1) + C ( x − 4)
2
x =1 → C = 43
2x 2
(x − 4)(x −1)
10.
2
and
x = 4 → A = 44
9
x = 0, A = 44
, C = 43 → B = 19
9
and
1
1
1
= 44 ⋅
+ 1⋅
+ 4⋅
3 ( x − 1)2
9 x −1
9 x−4
x2 − 2
A
Bx + C
=
+ 2
2
(x + 1) (x + 3) x + 1 x + 3
x = −1 → A = − 14
and
x 2 − 2 = A (x 2 + 3) + (Bx + C )( x + 1)
→
x = 0 → C = − 54
x = 1 → B = 54
and
x2 − 2
x −1
1
= − 1⋅
+ 5⋅ 2
2
4 x +1 4 x + 3
(x + 1) (x + 3)
11.
3x 2 + 9 x − 4
A
Bx + C
= + 2
2
(x −1) (x + 4 x −1) x x + 4 x −1
x =1 → A = 2
→
x=0 → C = 2
and
3x 2 + 9 x − 4 = A (x 2 + 4 x −1) + (Bx + C )( x − 1)
and
x = −1 → B = 1
3x 2 + 9 x − 4
2
x+2
=
+ 2
2
(x −1) (x + 4 x −1) x −1 x + 4 x −1
12.
(x − 5) (x
6
2
)
− 2x + 3
x = 5 → A = 13
=
A
Bx + C
+ 2
x − 5 x − 2x + 3
and
(
x = 0 → C = −1
)
6 = A x 2 − 2 x + 3 + (Bx + C )( x − 5)
→
and
x = 1 → B = − 13
x+3
1
= 1⋅
− 1⋅ 2
( x − 5) x − 2 x + 3 3 x − 5 3 x − 2 x + 3
(
6
2
)
Page 6 of 12
PART B:
Integration of proper Rational Expressions by Partial Fractions
In this part the student is expected to understand partial fraction decomposition as explained in
Part A. The student is also expected to be able to perform elementary integrations (by
substitution or by inspection) of the following types.
1
∫ ax + b dx
1
∫ (ax + b )
2
=
1
⋅ ln ax + b + C
a
1 1
dx = − ⋅
+ C
a ax + b
2ax + b
dx = ln ax 2 + bx + c + C
2
+ bx + c
∫ ax
In Part B each indefinite integral (antiderivative) must be simplified by decomposing the proper
rational expression into a sum of partial fractions. The details of the partial fraction
decomposition are left to the student. If necessary go back and review Part A.
Example 1
7 x +1
∫ (x + 3)(x −1) dx
Example 2
7 x +1
∫ (x + 3)(x −1) dx
Find the indefinite integral.
=
⎡ 5
∫ ⎢⎣ x + 3 +
2 ⎤
⎥ dx = 5 ln x + 3 + 2 ln x − 1 + C
x − 1⎦
∫
Find the indefinite integral.
6x 2 + 2x
dx
(x + 3)(x −1)(x − 5)
⎡
6x 2 + 2x
1
1
5 ⎤
∫ (x + 3)(x −1)(x − 5) dx = ∫ ⎢⎣ 32 ⋅ x + 3 − 12 ⋅ x −1 + x − 5 ⎥⎦ dx = 32 ln x − 3 − 12 ln x −1 + 5 ln x − 5 + C
Example 3
1− 3x
∫ x (x + 2)
2
Example 4
∫
Find the indefinite integral.
dx =
⎡
1
1
1
∫ ⎢ 14 ⋅ x − 14 ⋅ x + 2 − 72 ⋅ (x + 2)
⎣
Find the indefinite integral.
2
1 − 3x
dx
x + 4x 2 + 4x
3
⎤
7 1 +C
1
1
⎥ dx = 4 ln x − 4 ln x + 2 + 2 ⋅
x+2
⎦
x −3
∫ (x + 2)(x
2
+6
) dx
(continued next page)
Page 7 of 12
x −3
∫ (x + 2)(x
2
+6
Example 5
∫
)
⎡
1
∫ ⎢⎣− 12 ⋅ x + 2 + 12 ⋅ x
dx =
2
x ⎤
2
1
1
⎥ dx = − 2 ln x + 2 + 4 ln x + 6 + C
+ 6⎦
∫
Find the indefinite integral.
⎡
x 2 + 4x − 7
dx =
( x − 3) x 2 + 6 x − 6
(
1
∫ ⎢⎣ 23 ⋅ x − 3 + 13 ⋅ x
)
2
x 2 + 4x − 7
dx
(x − 3) x 2 + 6 x − 6
(
)
x+3 ⎤
2
2
1
⎥ dx = 3 ln x − 3 + 6 ln x + 6 x − 6 + C
+ 6x − 6 ⎦
EXERCISES Find the indefinite integral.
3 − 4x
dx
x2 + x
1.
∫
4.
3x 2 + 8x − 7
∫ (x + 4)(x + 3)(x + 1) dx
7.
∫
3 − 2x
dx
2
x + 6x + 9
5x 2 + 8x + 6
10. ∫
dx
(x + 4) (x 2 + 2)
x
dx
x + 7 + 10
2.
∫
5.
2 − 4x 2
∫ (x + 2)(x − 2)(x − 5) dx
2
3x −1
dx
3
x − 2x 2
8.
∫
11.
∫ (x + 3) (2 x
12 x + 18
dx
2
+ 8 x + 9)
6
dx
−14 x + 8
3.
∫ 3x
6.
∫ (x + 4)(x −1)(x − 3) dx
2
3x
2x 2 + x + 4
9.
∫ (x + 1)(x − 4)
12.
∫ (x − 4)(2 x
2
dx
15 − 25 x
dx
2
− 6x + 9
)
ANSWERS
3 − 4x
⎡3
1.
∫ x (x +1) dx
2.
∫ (x + 5)(x + 2) dx
3.
∫ (3x − 2)(x − 4) dx
4.
∫ ⎢⎣ x + 4 +
=
∫ ⎢⎣ x −
x
6
⎡ 3
=
=
7 ⎤
⎥ dx = 3 ln x − 7 ln x + 1 + C
x +1 ⎦
⎡
1 ⎤
1
∫ ⎢⎣ 53 ⋅ x + 5 − 23 ⋅ x + 2 ⎥⎦ dx
⎡
1
1 ⎤
= 53 ⋅ ln x + 5 − 23 ⋅ ln x + 2 + C
∫ ⎢⎣− 95 ⋅ 3x − 2 + 53 ⋅ x − 4 ⎥⎦ dx
= − 53 ⋅ ln 3 x − 2 + 53 ⋅ ln x − 4 + C
2
2 ⎤
−
⎥ dx = 3 ln x + 4 + 2 ln x + 3 − 2 ln x + 1 + C
x + 3 x + 1⎦
Page 8 of 12
⎡
1
1 ⎤
1
5.
∫ ⎢⎣− 12 ⋅ x + 2 + 76 ⋅ x − 2 − 143 ⋅ x − 5 ⎥⎦ dx
6.
dx
⋅
− 3⋅
+ 9⋅
∫ ⎢⎣− 12
35 x + 4 10 x − 1 14 x − 3 ⎥
⎦
⎡
1
3 − 2x
7.
∫ (x + 3)
8.
∫ x (x − 2) dx
9.
∫ ⎢ 15 ⋅ x +1
10.
∫ ⎢⎣ x + 4
11.
∫ ⎢⎣ x + 3
12.
∫ ⎢⎣ x − 4
2
3x −1
⎡
1
⎣
⎡ 3
⎡ −6
⎡ −5
∫
dx =
2
1 ⎤
1
=
= − 12 ⋅ ln x + 2 + 76 ⋅ ln x − 2 − 14
⋅ ln x − 5 + C
3
3 ⋅ ln x − 1 + 9 ⋅ ln x − 3 + C
= − 12
⋅ ln x + 4 − 10
35
14
⎡ −2
9 ⎤
9
−
dx = − 2 ln 3x − 2 +
+ C
⎢
2 ⎥
x+3
⎣ x + 3 ( x + 3) ⎦
⎡
1
∫ ⎢⎣− 54 ⋅ x
1
1 ⎤
5
5
1 1
+ 12 ⋅ 2 + 54 ⋅
⎥ dx = − 4 ⋅ ln x − 2 ⋅ + 4 ⋅ ln x − 2 + C
x − 2⎦
x
x
1
8 ⎤
8
+ 95 ⋅
+
dx = 15 ⋅ ln x + 1 + 95 ⋅ ln x − 4 −
+ C
2 ⎥
x−4
x−4
(x − 4) ⎦
+
2x ⎤
2
⎥ dx = 3 ln x + 4 + ln x + 2 + C
x + 2⎦
+
12 x + 24 ⎤
2
⎥ dx = − 6 ln x + 3 + 3 ln 2 x + 8 x + 9 + C
2
2 x + 8x + 9 ⎦
+
10 x −15 ⎤
2
5
⎥ dx = − 5 ln x − 4 + 2 ln 2 x − 6 x + 9 + C
2x 2 − 6x + 9 ⎦
2
PART C:
Integration of improper Rational Expressions by Partial Fractions
The prerequisite skills of Part B are also required in Part C. Furthermore, the student is expected
to be able to use long division to decompose an improper rational expression.
In each indefinite integral of Part C the improper rational expression must be rewritten as a
polynomial plus proper rational. In most cases the resulting proper rational can then be further
simplified by decomposition into a sum of partial fractions. The details of the long division and
partial fraction decomposition are left to the student.
Example 1
∫
Find the indefinite integral.
3 x 3 + 8 x 2 −10 x + 15
dx =
x+4
⎡
∫ ⎢⎣3x
2
− 4x + 6 −
∫
3x 3 + 8 x 2 − 10 x + 15
dx
x+4
9 ⎤
3
2
⎥ dx = x − 2 x + 6 x − 9 ln x + 4 + C
x + 4⎦
Page 9 of 12
Example 2
∫
2x 2 + 4x − 7
dx =
x2 + x −6
→
∫
+
∫
∫ ⎢2 x
⎣
1
1 ⎤
+ 15 ⋅
+ 95 ⋅
⎥ dx
x+3
x − 2⎦
⎡
∫
2 x 3 + 2 x 2 − 95 x + 40
dx
x 3 + x 2 − 20 x
⎤
40 −11x
∫ ⎢⎣2 + x(x + 5)(x − 4)⎥⎦ dx
Find the indefinite integral.
2 x 4 − 9x 3 + 9x 2 − 8x − 6
dx =
x3 − 6x 2 + 9x
⎡
∫ ⎢⎣2
=
⎡
∫ ⎢⎣2
−
2
7
5 ⎤
+
−
⎥ dx
x
x+5
x − 4⎦
2 x 3 + 2 x 2 − 95 x + 40
dx = 2 x − 2 ln x + 7 ln x + 5 − 5 ln x − 4 + C
x 3 + x 2 − 20 x
Example 4
=
⎡
2x + 5 ⎤
dx =
(x + 3)(x − 2)⎥⎦
Find the indefinite integral.
2 x 3 + 2 x 2 − 95 x + 40
dx =
x 3 + x 2 − 20 x
→
∫
⎡
∫ ⎢⎣2
∫
2x 2 + 4x − 7
dx
x2 + x − 6
2x 2 + 4x − 7
dx = 2 x + 15 ⋅ ln x + 3 + 95 ⋅ ln x − 2 + C
x2 + x − 6
Example 3
∫
Find the indefinite integral.
∫
∫
2 x 4 − 9 x 3 + 9 x 2 − 8x − 6
dx
x3 − 6x 2 + 9x
⎡
9 x 2 − 35 x − 6 ⎤
⎢2 x + 3 +
⎥ dx
2
x ( x − 3) ⎦
⎣
1
1
10 ⎤
+ 3 − 23 ⋅ + 29
−
⋅
⎥ dx
3 x −3
x
(x − 3)2 ⎦
10
= x 2 + 3x − 23 ⋅ ln x + 29
⋅ ln x − 3 +
+ C
3
x −3
EXERCISES Find the indefinite integral.
1.
4.
7.
∫
6x + 5
dx
x+2
∫
x 3 + 3x 2 − 4 x − 6
dx
x 2 + 2 x −15
∫
4 x 3 + 20 x 2 + 15 x + 8
dx
x 3 + 5x 2 + 4x
2.
5.
∫
4 x 2 − 12 x − 25
dx
x −5
∫
4 x 2 − 8x + 3
dx
x 2 − 3x − 4
8.
∫
3.
6.
∫
5x 3 + 3x − 2
dx
x −1
∫
x 3 − 3x 2
dx
x 2 − 3 x − 10
x 4 − 10 x 3 + 28 x 2 − 15 x − 15
dx
x 3 − 7 x 2 + 10 x
Page 10 of 12
9.
∫
x 5 − 2 x 4 − 7 x 3 + 20 x 2 −12 x + 4
dx
x3 + x 2 − 6x
∫
11.
10.
2 x 3 − 15 x 2 + 17 x + 25
dx
x 3 −10 x 2 + 25 x
12.
∫
4 x 5 + 6 x 4 + 2 x 3 + 3x 2 − 5 x − 7
dx
x3 + 2x 2 + x
∫
x 4 + 2x3 − 6x 2 − 6x + 3
dx
x3 + 4x 2 + 4x
ANSWERS
⎡
7 ⎤
⎥ dx = 6 x − 7 ln x + 2 + C
x + 2⎦
∫
6x + 5
dx =
x+2
∫
4 x 2 − 12 x − 25
dx =
x −5
∫ ⎢⎣4 x + 8 +
3.
∫
5 x 3 + 3x − 2
dx =
x −1
⎡
4. ∫
x 3 + 3x 2 − 4 x − 6
dx =
x 2 + 2 x − 15
1.
2.
=
5.
∫
⎡
∫ ⎢⎣ x + 1 +
∫ ⎢⎣6 −
⎡
∫ ⎢⎣5 x
2
15 ⎤
2
⎥ dx = 2 x + 8 x + 15 ln x − 5 + C
x −5⎦
+ 5x + 8 +
⎡
6 ⎤
5 3
5 2
⎥ dx = 3 x + 2 x + 8 x + 6 ln x − 1 + C
x − 1⎦
⎤
9x + 9
∫ ⎢⎣ x + 1 + (x + 5)(x − 3)⎥⎦ dx
9 ⋅ 1 + 9 ⋅ 1 ⎤ dx = 1 x 2 + x + 9 ⋅ ln x + 5 + 9 ⋅ ln x − 3 + C
2 x+5
2 x − 3⎥
2
2
2
⎦
4 x 2 − 8x + 3
dx =
x 2 − 3x − 4
⎡
4 x + 19
⎤
∫ ⎢⎣4 + (x + 1)(x − 4)⎥⎦ dx
⎡
=
∫ ⎢⎣4 −
=
∫ ⎢⎣ x
3
7 ⎤
+
⎥ dx
x +1 x − 4 ⎦
= 4 x − 3 ln x + 1 + 7 ln x − 4 + C
6.
∫
x 3 − 3x 2
dx =
x 2 − 3 x − 10
⎡
10 x
⎤
∫ ⎢⎣ x + (x + 2)(x − 5)⎥⎦ dx
⎡
1
50 ⋅ 1 ⎤ dx
⋅
+ 20
+
7 x+2
7 x −5⎥
⎦
50
= 12 x 2 + 20
7 ⋅ ln x + 2 + 7 ⋅ ln x − 5 + C
7.
∫
4 x 3 + 20 x 2 + 15 x + 8
dx =
x 3 + 5x 2 + 4x
⎡
8− x
⎤
∫ ⎢⎣4 + x (x + 4)(x + 1)⎥⎦ dx
=
⎡
∫ ⎢⎣4 +
2
1
3 ⎤
+
−
⎥ dx
x x + 4 x + 1⎦
= 4 x + 2 ln x + ln x + 4 − 3 ln x + 1 + C
Page 11 of 12
8.
∫
x 4 − 10 x 3 + 28 x 2 − 15 x −15
dx =
x 3 − 7 x 2 + 10 x
=
9.
∫
⎡
1
1
∫ ⎢⎣ x − 3 − 32 ⋅ x − 12 ⋅ x − 2 −
∫
⎡
− 3 x 2 + 15 x − 15 ⎤
x
−
3
+
⎢
⎥ dx
x ( x − 2 )( x − 5) ⎦
⎣
1 ⎤
3
1 2
1
⎥ dx = 2 x − 3 x − 2 ln x − 2 ln x − 2 − ln x − 5 + C
x −5⎦
⎡
x 5 − 2 x 4 − 7 x 3 + 20 x 2 −12 x + 4
dx =
x3 + x 2 − 6x
=
⎡
∫ ⎢⎣ x
2
∫ ⎢⎣ x
2
− 3x + 2 +
⎤
4
⎥ dx
x ( x + 3)(x − 2 ) ⎦
1
4 ⋅ 1 + 2 ⋅ 1 ⎤ dx
− 3 x + 2 − 23 ⋅ + 15
5 x − 2⎥
x
x+3
⎦
4 ln x + 3 + 2 ln x − 2 + C
= 13 x 3 − 32 x 2 + 2 x − 23 ln x + 15
5
10. ∫
4 x 5 + 6 x 4 + 2 x 3 + 3x 2 − 5x − 7
dx =
x3 + 2x 2 + x
=
∫
∫
⎡ 2
x2 − 7x − 7 ⎤
x
−
x
+
+
dx
4
2
2
⎢
2 ⎥
x ( x + 1) ⎦
⎣
⎡ 2
7
8
1 ⎤
+
−
⎢4 x − 2 x + 2 −
⎥ dx
x
x +1
(x + 1)2 ⎦
⎣
= 43 x 3 − x 2 + 2 x − 7 ln x + 8 ln x + 1 +
2 x 3 −15 x 2 + 17 x + 25
dx =
11. ∫
x 3 − 10 x 2 + 25 x
∫
⎡
5 x 2 − 33 x + 25 ⎤
⎢2 +
⎥ dx =
2
x ( x − 5)
⎣
⎦
= 2 x + ln x + 4 ln x − 5 −
12. ∫
x 4 + 2x3 − 6x 2 − 6x + 3
dx =
x3 + 4x 2 + 4x
=
⎡
∫ ⎢x
⎣
1
+ C
x +1
∫
∫
⎡
1
4
3 ⎤
−
⎢2 + +
⎥ dx
x x − 5 ( x − 5 )2 ⎦
⎣
3
+ C
x −5
⎡
− 2x 2 + 2x + 3⎤
⎢x − 2 +
⎥ dx
2
x (x + 2) ⎦
⎣
1
1
1 ⎤
9⋅
− 2 + 34 ⋅ − 11
+
⋅
dx
4 x+2
2 ( x + 2 )2 ⎥
x
⎦
1
= 12 x 2 − 2 x + 34 ⋅ ln x − 11
⋅ ln x + 1 − 9 ⋅
+ C
2 x+2
4
Page 12 of 12