CHAPTER 2
PRIMITIVES AND INTEGRALS
Mathematical Analysis 1 , ENSIA 2024
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PRIMITIVES AND INTEGRALS
Part III
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Antiderivatives of a Rational Function
1
Polynomial Reminders
2
Decomposition of a Rational Function:
3
Antiderivatives of a Rational Function
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Polynomial Reminders
Definition
A polynomial in one complex variable z is any expression of the form:
P (z ) = a0 + a1 z + a2 z 2 + . . . + an z n
Where: a0 , a1 , a2 , . . . , an ∈ C are called the coefficients of the polynomial.
If an 6= 0, we say that the polynomial is of degree n.
The derivative of the polynomial P, denoted as P 0 , is defined as:
P 0 (z ) = a1 + 2a2 z + . . . + nan z (n−1)
This represents the derivative of P with respect to z.
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Polynomial Reminders
Roots of a Polynomial
We say that λ ∈ C is a root of the polynomial P if P (λ) = 0. This
definition is equivalent to P being divisible by (z − λ), meaning there
exists a polynomial Q such that P (z ) = (z − λ)Q (z ).
The roots of a polynomial can be either simple or multiple.
We say that λ ∈ C is a simple root if:
(
P (λ) = 0
P (z ) = (z − λ)Q (z ), Q (λ) 6= 0 =⇒
P 0 (λ) 6= 0
We say that λ ∈ C is a double root if:
(
P (z ) = (z − λ )2 Q (z ),
Q (λ) 6= 0 =⇒
P (λ) = P 0 (λ) = 0
P 00 (λ) 6= 0
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Polynomial Reminders
**More generally:**
We will say that λ ∈ C is a root of multiplicity or order k ∈ N∗ if:
P (z ) = (z − λ )k Q (z ),
(
=⇒
Q (λ) 6= 0
P ( λ ) = P 0 ( λ ) = . . . = P (k −1) ( λ ) = 0
P (k ) ( λ ) 6 = 0
Example
P (z ) = z 3 − 1
There are 3 simple roots: z1 = 1, z2 = −1−2i
√
3
, and z3 = −1+2i
√
3
.
Example
P (z ) = z 5 − 3z 4 + 4z 3 − 4z 2 + 3z − 1
z1 = 1 is a triple root because P (1) = P 0 (1) = P 00 (1) = 0, and
000
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Example
P (z ) = z 4 + 4z 3 + mz 2 + nz + 2 where m, n ∈ N
Choose m and n such that −1 is a double root.
We need to have P (−1) = P 0 (−1) = 0 and P 00 (−1) 6= 0.
P 0 (z ) = 4z 3 + 12z 2 + 2mz + n
P 00 (z ) = 12z 2 + 24z + 2m
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Example
Setting z = −1:
P 0 (−1) = 4(−1)3 + 12(−1)2 + 2m (−1) + n = 0
P 00 (−1) = 12(−1)2 + 24(−1) + 2m 6= 0
Solving the equations:
m−n−1 = 0
8 − 2m + n = 0
This simplifies to:
m=7
n=6
So, m = 7 and n = 6 satisfy the conditions.
Also, −1 + i and −1 − i are also roots of P.
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Polynomial Reminders
Case of Polynomials with Real Coefficients
Let n ∈ N∗ and P (z ) = a0 + a1 z + a2 z 2 + . . . + an z n where
a0 , a1 , a2 , . . . , an ∈ R.
If λ ∈ C is a root of the polynomial, then the conjugate λ is also a root.
**Proof:**
P (λ) = a0 + a1 λ + a2 λ2 + . . . + an λn = 0
Taking the conjugate of both sides:
P (λ) = a0 + a1 λ + a2 λ2 + . . . + an λn = 0
2
n
P (λ) = a0 + a1 λ + a2 λ + . . . + an λ = P (λ) = 0
As a consequence, if λ ∈ C is a root of order k ≥ 1, then the conjugate λ
is also a root of order k of the polynomial.
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Polynomial Reminders
Polynomial Factorization
Consider P (z ) = a0 + a1 z + a2 z 2 + . . . + an z n , where
a0 , a1 , a2 , . . . , an ∈ C, a polynomial of degree n ≥ 1. It is assumed that it
has at least one complex root, i.e., ∃λ ∈ C such that P (λ) = 0.
**Note:** This result is not valid in R.
For example, consider P1 (x ) = x 2 + x + 1 and P2 (x ) = x 4 + 1.
The polynomials P1 (x ) and P2 (x ) do not have real roots. However, in C,
P1 (x ) has complex roots, and P2 (x ) has complex roots, even though they
are not real.
This emphasizes that, unlike in R, in C, every non-constant polynomial
has at least one complex root.
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Polynomial Reminders
**Consequences:**
If P (z ) = a0 + a1 z + a2 z 2 + . . . + an z n where a0 , a1 , a2 , . . . , an ∈ C is a
polynomial of degree n ≥ 1, then it has n distinct or non-distinct roots.
If λ1 , λ2 , . . . , λp are the distinct roots with respective orders k1 , k2 , . . . , kp
(where k1 + k2 + . . . + kp = n), then:
P (z ) = an (z − λ1 )k1 (z − λ2 )k2 . . . (z − λp )kp
Now, if P (z ) = a0 + a1 z + a2 z 2 + . . . + an z n where a0 , a1 , a2 , . . . , an ∈ R
is a polynomial of degree n ≥ 1, and λ ∈ C is a root of multiplicity k,
then the conjugate λ is also a root.
In the expression of P, you find the factor (z − λ)k (z − λ)k .
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Polynomial Reminders
**Particularly, in the case where P is real, i.e., of the form:**
P (x ) = a0 + a1 x + a2 x 2 + . . . + an x n
where a0 , a1 , a2 , . . . , an ∈ R, x ∈ R,
Then, the factorization of such a polynomial in R gives us factors of the
form (x − a)k or (x 2 + px + q )m with ∆ < 0.
Example
P (x ) = x 4 + 1
√
√
P (x ) = (x 2 + 1)2 − 2x 2 = (x 2 − x 2 + 1)(x 2 + x 2 + 1)
In this example, the factorization involves quadratic factors with ∆ < 0.
the factorization does not depend on the root it can be done only by famous
identification
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Polynomial Reminders
Example
P (x ) = x 5 + 1
Let’s find the roots in C. Suppose z is such a root.
z = ρe i θ =⇒ z 5 = ρ5 e 5i θ .
(
5
=⇒
ρ=1
θ = π5 + 2πk
5 ,
k = 0, 1, 2, 3, 4
Z2 = (Z1 )∗ = e −i π/5 ,
Z3 = e i (3π/5) ,
Z4 = (Z3 )∗ = e −i3π/
5 5i θ
z = −1 =⇒ ρ e
=e
iπ
So, we have 5 roots:
Z1 = e i π/5 ,
Therefore, the factorization of P (x ) is:
P (x ) = (x + 1)(x 2 − 2x cos(π/5) + 1)(x 2 − 2x cos(3π/5) + 1)
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Decomposition of a Rational Function
Fraction Rationale
Consider P and Q, two real polynomials that are coprime (have no
common factors).
P (x )
The function F : x 7→ F (x ) = Q (x ) , where Q (x ) 6= 0, is called a rational
function. F is said to be proper if the degree of P is strictly less than the
degree of Q (deg(P ) < deg(Q )).
The roots of Q are called the ”poles” of F . A root of order k of the
denominator is called a ”pole” of order k.
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Decomposition of a Rational Function
Example
F (x ) =
x2 + 1
x (x − 1)2 (x + 1)3
x = 0 is a simple pole
x = 1 is a double pole
x = −1 is a triple pole
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Decomposition of a Rational Function
Definition
A partial (or elementary) rational fraction is a fraction of the form:
A
(x − a )k
or
Cx + D
(x 2 + px + q )j
where k, j ∈ N∗ and ∆ = p 2 − 4q < 0.
P (x )
Let F (x ) = Q (x ) be a proper rational fraction. We know that the
denominator can be factored into factors of the form (x − a)k or
(x 2 + px + q )j where ∆ < 0. The former are called simple elements of the
first kind, and the latter are simple elements of the second kind.
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Decomposition of a Rational Function
Theorem
Every proper rational fraction can be uniquely expressed as the sum of
partial rational fractions. For simplicity, the procedure is outlined as
follows:
For each simple element of the first kind (x − a)k in the partial fraction
decomposition, there should be a sum of the form:
Ak −1
A1
Ak
+
+...+
(x − a )
(x − a )k (x − a )k −1
For each simple element of the second kind (x 2 + px + q )j with ∆ < 0 in
the partial fraction decomposition, there should be a sum of the form:
Cj x + Dj
C1 x + D 1
C2 x + D 2
+ 2
+...+ 2
2
2
x + px + q (x + px + q )
(x + px + q )j
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Decomposition of a Rational Function
Example
F (x ) =
F (x ) =
x2 + 1
x 2 (x − 1)(x + 1)3
A2 A1
B1
C3
C2
C1
+
+
+
+
+
2
3
2
x
x
(x − 1) (x + 1)
(x + 1)
(x + 1)
Example
F (x ) =
F (x ) =
x2 + x + 1
(x − 1)2 (x − 2)3 (x 2 + 1)2
A2
A1
B3
+
+
+
2
(x − 1)
(x − 1) (x − 2)3
B2
D1
· (x −
+ (xB−12) + C(1xx2 +
+ C(x22x++1D)22
2)2
+1)
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Decomposition of a Rational Function
Calculating Coefficients
**Method of Undetermined Coefficients:** In the decomposition of F into
partial fractions, you multiply both sides by the denominator, expand, and
then identify the coefficients of the monomials of the same power.
Example
Decompose the rational function into partial fractions:
F (x ) =
We have:
F (x ) =
x2 + 1
x (x − 1)(x + 2)
A
B
C
+
+
x
(x − 1) (x + 2)
Now, you would proceed to find the values of A, B, and C .
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Decomposition of a Rational Function
Example
(continued)
x 2 + 1 = (A + B + C )x 2 + (A + 2B − C )x − 2A
By identification, we get the system of equations:
−2A = 1
A + 2B − C = 0
A+B +C = 1
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Decomposition of a Rational Function
Example
(continued) Solving this system:
A=−
1
2
2
3
5
C =
6
Therefore, the partial fraction decomposition is:
B=
x2 + 1
1
2
5
=− +
+
x (x − 1)(x + 2)
2x
3(x − 1) 6(x + 2)
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Decomposition of a Rational Function
**Method of Division According to Increasing Powers:**
The previous method can be quite lengthy. Here’s an alternative method
for calculating coefficients for the pole of order k, k ≥ 1.
Suppose the denominator is Q (x ) = (x − a)k Q1 (x ) where Q1 (a) 6= 0,
meaning that a is a pole of order k.
The partial fraction for this pole is of the form:
Ak −1
A1
Ak
+
+...+
k
k
−
1
(x − a )
(x − a )
(x − a )
Definition
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Decomposition of a Rational Function
**Two Cases:** 1. **k = 1:** There is only one coefficient to calculate,
A1 :
P (a )
A1 = lim (x − a)F (x ) =
x →a
Q (a )
2. **k > 1:** In the rational fraction P (x )/Q1 (x ) = (x − a)k F (x ), set
x = a + y and perform division according to increasing powers of y up to
order k − 1.
Example
F (x ) =
F (x ) =
x2 + 1
x 2 (x − 1)(x + 1)3
A2 A1
B1
C3
C2
C1
+
+
+
+
+
2
3
2
x
x
(x − 1) (x + 1)
(x + 1)
(x + 1)
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Decomposition of a Rational Function
Example
(continued):**
F (x ) =
F (x ) =
x2 + 1
x 2 (x − 1)(x + 1)3
C2
C1
A2 A1
B1
C3
+
+
+
+
+
2
3
2
x
x
(x − 1) (x + 1)
(x + 1)
(x + 1)
For the pole at x = 1:
x2 + 1
1
=
2
3
x →1 x (x + 1 )
4
B1 = lim (x − 1)F (x ) = lim
x →1
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Decomposition of a Rational Function
Example
(continued):** Since x = 0 is a double pole, we perform division according
to increasing powers up to order 1:
F (x ) =
1 + x2
1 + x2
1 + x2
=
=
(x − 1)(x + 1)3
(x − 1)(1 + 3x + . . .)
(−1 − 2x + . . .)
= -1 + 2x + . . .
This implies A2 = −1 and A1 = 2.
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Decomposition of a Rational Function
Example
(continued)
F (x ) =
x2 + 1
x 2 (x − 1)(x + 1)3
For the triple pole at x = −1, we set x + 1 = y , meaning x = −1 + y ,
and then perform division according to increasing powers up to order 2:
x2 + 1
(−1 + y )2 + 1
2 − 2y + y 2
=
=
x 2 (x − 1)
(−1 + y )2 (−2 + y )
−2 + 5y − 4y 2 + . . .
This implies C3 = −1, C2 = − 23 , and C1 = − 49 .
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Decomposition of a Rational Function
Example
F (x ) =
x3 − 1
x 3 (x + 1)2 (x 2 + 1)
For the triple pole at x = 0:
F (x ) =
A3 A2 A1
B1
B2
C1 x + D 1
+ 2 +
+
+
+ 2
3
2
x
x
x
(x + 1) (x + 1)
x +1
We perform division according to increasing powers up to order 2 of
x 3 F (x ):
x 3 F (x ) =
−1 + x 3
−1 + x 3
2
(
1
+
x
)
=
1 + 2x + x 2
1 + 2x + 2x 2 + . . .
This implies A3 = −1, A2 = 2, and A1 = −2.
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Decomposition of a Rational Function
Example
(continued):
(x + 1)2 F (x ) =
x3 − 1
(−1 + y )3 − 1
−2 + 3y + . . .
=
=
3
2
3
2
x (x + 1)
(−1 + y ) ((−1 + y ) + 1)
−2 + 8y + . . .
This implies B2 = 1 and B1 = 25 .
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Decomposition of a Rational Function
Example
(continued): We have two coefficients left to determine. We can either
give x two values other than 0 and -1, or we can use:
lim xF (x ) = 0 = A1 + B1 + C1
x →+∞
From this, we find that C1 = −A1 − B1 = 2 − 25 = − 12 .
To find D1 , we give x the value 1.
F (1) = 0 = A3 + A2 + A1 +
B1 B2 C1
+
+
+ D1
2
4
2
Solving for D1 , we get D1 = − 34 .
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Decomposition of a Rational Function
Example
F (x ) =
1
(x 2 − 1)(x 2 + 1)
We have:
F (x ) =
A
B
Cx + D
1
=
+
+ 2
(x − 1)(x + 1)(x 2 + 1)
x −1 x +1
x +1
Due to the parity of the rational function, we know that F (−x ) = F (x ),
which implies B = −A and C = 0.
A
A
D
Therefore, F (x ) = x −
1 − x +1 + x 2 +1 .
By finding the limit as x approaches 1, we get A = 41 .
To find D, we give x the value 0.
1
F (0) = −1 = −2A + D = − + D
2
Solving for D, we find D = − 21 .
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Decomposition of a Rational Function
** general case**
P (x )
Let F (x ) = Q (x ) be a rational function, whether proper or not.
If deg(P ) < deg(Q ), we directly proceed to partial fraction decomposition.
If deg(P ) > deg(Q ), we perform either polynomial long division or division
following the decreasing powers of P (x ) by Q (x ). This results in two
unique polynomials E (x ) (quotient) and R (x ) (remainder) such that
P (x ) = E (x )Q (x ) + R (x ), where deg(R ) < deg(Q ).
R (x )
P (x )
This implies F (x ) = Q (x ) = E (x ) + Q (x ) = E (x ) + F1 (x ), where F1 (x ) is
proper and can be expressed as a sum of partial fractions.
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Decomposition of a Rational Function
Example
f (x ) =
5x 2 + x − 5
x4 + x − 1
=
x
−
2
+
x 3 + 2x 2 − x − 2
(x + 2)(x 2 − 1)
2
+x −5
A
We have F1 (x ) = (x 35x
= (x +
+ (x B−1) + (x C+1) . Solving for A,
+2x 2 −x −2)
2)
B, and C using the limits:
13
x →−2
3
1
B = lim (x − 1)F1 (x ) =
x →1
6
1
C = lim (x + 1)F1 (x ) =
x →−1
2
A = lim (x + 2)F1 (x ) =
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Decomposition of a Rational Function
Example
Thus, the decomposition is:
f (x ) = x − 2 +
1
1
13
+
+
3(x + 2) 6(x − 1) 2(x + 1)
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Antiderivatives of a Rational Function
P (x )
The decomposition of F (x ) = Q (x ) generally results in a polynomial part
E (x ) and partial fractions of the form:
A
Cx + D
or 2
k
(x + px + q )j
(x − a )
where k, j ∈ N∗ and ∆ = p 2 − 4q < 0.
If E (x ) = a0 + a1 x + . . . + ai x i ,
Z
E (x ) dx = a0 x +
ai i +1
a1 2
x +...+
x
+C
2
i +1
Similarly,
Z
1
dx =
(x − a )k
(
ln |x − a| + C
1
1−k + C
1−k (x − a )
if k = 1
if k 6= 1
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Antiderivatives of a Rational Function
Ax +B
Therefore, we need to find the antiderivatives of (x 2 +
. We have:
px +q )m
Z
A
=
2
Z
Ax + B
dx =
(x 2 + px + q )m
Z A
A
2 (2x + p ) + B − 2 p
(x 2 + px + q )m
2x + p
A
dx + (B − p )
(x 2 + p + q )m
2
Z
dx
1
dx
(x 2 + px + q )m
2x + p
ϕ 0 (x )
dx
=
dx
(x 2 + px + q )m
ϕ (x )m
(
Z
ln |t | + C
if m = 1
dt
=
=
m
1
1
−
m
t
+ C if m 6= 1
1−m t
Z
Z
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Antiderivatives of a Rational Function
Therefore:
2x + p
Z
(x 2 + px + q )m
(
=
ln |x 2 + px + q | + C
1
2
1−m + C
1−m (x + px + q )
dx
if m = 1
if m 6= 1
Now, for the antiderivatives of (x 2 +px1 +q )m :
p
p2
x 2 + px + q = (x + )2 + (q − )
2
4
p
x 2 + px + q = (x + )2 + α2
2
where α =
q
2
q − p4 .
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Antiderivatives of a Rational Function
This expression can also be written as:
x 2 + px + q = α2 (1 + (
Z
1
dx =
2
(x + px + q )m
Z
x + p2 2
) )
α
1
α2m (1 + (
x + p2 2 m
α ) )
dx
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Antiderivatives of a Rational Function
x+ p
Let’s perform the substitution α 2 = t, which implies x = αt − p2 . Then,
calculate the differential dx in terms of dt:
dx = αdt
Now, substitute x and dx in the integral:
Z
=α
Now, define Im =
R
1
dx
(x 2 + px + q )m
1−2m
Z
1
dt
(1 + t 2 )m
1
dt. Rewrite the integral in terms of Im :
1 (1+t 2 )m
α
1−2m
Z
1
dt = α2m−1 Im
(1 + t 2 )m
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Antiderivatives of a Rational Function
for m = 1
I1 =
Z
1
dt
1 + t2
I1 = arctan(t ) + C
Let’s evaluate Im+1 for m > 1 step by step.
Given:
Z
1 + t2 − t2
dt
Im+1 =
(1 + t 2 )m +1
Im+1 =
Z
1
1
dt −
2
m
(1 + t )
2
Z
t
2t
dt
(1 + t 2 )m +1
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Antiderivatives of a Rational Function
Let’s perform integration by parts with the chosen u and dv :
u = t =⇒ du = dt
1
−1
2t
dt =⇒ v =
dv =
2
m
+
1
(1 + t )
m (1 + t 2 )m
Now, apply the integration by parts formula:
1 −1
t
1
Im+1 = Im − (
+
2
m
2 m (1 + t )
m
Im+1 = Im (1 −
Im+1 =
Z
1
dt )
(1 + t 2 )m
1
1
t
)+
2m
2m (1 + t 2 )m
2m − 1
1
t
Im +
2m
2m (1 + t 2 )m
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Antiderivatives of a Rational Function
Example
3
Primitives of f (x ) = x 3 (x +x1)−2 (1x 2 +1) are obtained by decomposing it into
partial fractions, as seen in Example 2. In this regard, the expression for
f (x ) is as follows:
1
5/2
1
2
2
1/2x + 3/4
f (x ) = − 3 + 2 − +
+
−
x
x
x
(x + 1) (x + 1)2
(x 2 + 1)
The indefinite integral of f (x ) is given by:
Z
R
f (x ) dx =
2
5
1
1
1
− − 2 ln(x ) + ln(1 + x ) −
−
2x 2 x
2
x +1 4
Z
2x
3
dx −
x2 + 1
4
1
dx
x 2 +1
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Antiderivatives of a Rational Function
Example
(continued)
Z
f (x ) dx =
1
2
5
1
1
3
− − 2 ln(x ) + ln(1 + x ) −
− ln(x 2 + 1) −
2x 2 x
2
x +1 4
4
arctan(x ) + C
where C is the constant of integration.
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Antiderivatives of a Rational Function
Example
2
Decomposition and indefinite integral of f (x ) = (x −1)2 (xx−2+)(1x 2 +x +1)
In this case, f (x ) can be expressed as the sum of partial fractions:
f (x ) =
B
Cx + D
A2
A1
+
+
+
(x − 1)2 (x − 1) (x − 2) x 2 + x + 1
Calculation of coefficients:
B = lim (x − 2)f (x ) =
x →2
(x − 1)2 f (x ) =
5
7
1 + x2
2 2
=− − y
2
(x − 2)(x + x + 1)
3 3
where x =y+1.
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Antiderivatives of a Rational Function
Example
A1 = A2 = −
2
3
lim xf (x ) = 0 = A1 + B + C =⇒ C = −A1 − B = −
x →∞
f (0) = −
1
21
1
1
1
1
1
= A2 − A1 − B + D =⇒ D = − (−A2 + A1 + B ) = −
2
2
2
2
7
1
Therefore, the coefficients are A1 = A2 = − 23 , B = 57 , C = − 21
, and
1
D = −7.
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Antiderivatives of a Rational Function
Example
2
Example 2 (continued): Indefinite Integrals of f (x ) = (x −1)2 (xx−2+)(1x 2 +x +1)
In this case, the partial fraction decomposition of f (x ) was found to be:
2
1
1
5
1
2x + 1
5
2
f (x ) = − ·
+ ·
−
−
− ·
3 (x − 1)2 3 (x − 1) 7 (x − 2) 42(x 2 + x + 1) 42
· (x 2 +1x +1)
This expression can be simplified further:
2
1
2
1
5
1
1
2x + 1
5
f (x ) = − ·
− ·
+ ·
− ·
−
3 (x − 1)2 3 (x − 1) 7 (x − 2) 42 (x 2 + x + 1) 42
· (x 2 +1x +1)
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Antiderivatives of a Rational Function
Example
(continued):
Z
R
Z
R
f (x ) dx = −
Z
2
dx −
3(x − 1)2
Z
2
dx +
3(x − 1)
Z
5
1
dx −
7(x − 2)
42
R
(2x +1)
5
dx − 42(x 2 +
dx
(x 2 +x +1)
x +1)
f (x ) dx =
2
2
5
1
− ln |x − 1| + ln |x − 2| −
ln(x 2 + x + 1)−
3(x − 1) 3
7
42
5
dx
42(x 2 +x +1)
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Antiderivatives of a Rational Function
Example
(continued):
2
x +x +1 =
Z
2
2 !
1+
x + 12
4
1
dx =
2
x +x +1
3
1
x+
2
3
3
+ =
4
4
3
4
1
Z
1+
2x√+1
3
dx =
1+
2x + 1
√
3
2 !
2 dx
We make the substitution t = 2x√+31 , which implies x = t
√
3
=
4
√
3−1
and
2
3
2 dt.
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Antiderivatives of a Rational Function
Example
(continued):
Z
1
2
1
2
5
dx =
− ln |x − 1| + ln |x − 2|
x2 + x + 1
3 (x − 1) 3
7
5
1
2x + 1
2
√
− ln(x + x + 1) − √ arctan
+C
42
21 3
3
So, the final result is:
Z
1
2
x2 + 1
2
5
dx =
− ln |x − 1| + ln |x − 2|
2
2
(x − 1) (x − 2)(x + x + 1)
3 (x − 1) 3
7
1
5
2x + 1
2
√
− ln(x + x + 1) − √ arctan
+C
42
21 3
3
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Antiderivatives of a Rational Function
Example
f (x ) =
f (x ) =
x2 + 2
(x − 1)(x 2 + 1)2
B1 x + C1
B2 x + C2
A
+ 2
+ 2
(x − 1)
(x + 1)
(x + 1)2
A = lim (x − 1)f (x ) =
x →1
3
4
1
1
lim (x 2 + 1)2 f (x ) = − (1 + i ) = B2 i + C2 =⇒ B2 = C2 = −
x →i
2
2
lim xf (x ) = 0 = A + B1 =⇒ B1 = −
x →∞
3
4
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Antiderivatives of a Rational Function
Example
(continued):
f (0) = −2 = −A + C1 + C2
=⇒ C1 = −2 + A − C2 = −
3
4
Hence:
f (x ) =
1
3
2x
3
1
1
2x
1
3
−
·
− ·
− ·
− ·
4 (x − 1) 8 (x 2 + 1) 4 (x 2 + 1) 4 (x 2 + 1)2 2
· (x 2 +1 1)2
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Antiderivatives of a Rational Function
Example
(continued):
Z
f (x ) dx =
3
3
3
1 1
ln |x − 1| − ln(x 2 + 1) − arctan(x ) +
4
8
4
4 1 + x2
−
To compute
R
R
Z
1
dx
2 + 1)2
(
x
1
1
dx, we apply the formula:
(x 2 +1)2
Im+1 =
Where Im =
1
2
1
t
2m − 1
Im +
2m
2m (1 + t 2 )m
1
dx.
(x 2 +1)m
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Antiderivatives of a Rational Function
Example
(continued):
I2 =
1 x
1
I1 +
2
2 1 + x2
1
1
I1 = arctan(x ) + C
2
2
In conclusion, we have:
Z
f (x ) dx =
3
3
3
1 1
ln |x − 1| − ln(x 2 + 1) − arctan(x ) +
4
8
4
4 1 + x2
1
1 x
− arctan(x ) −
+K
4
4 1 + x2
where K is the constant of integration.
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