7.

4 Integration of rational functions by partial fractions

Let’s consider a rational function
P (x)
f (x) =
Q(x)
where P and Q are polynomials.

1st step
If deg(P ) ≥ deg(Q), then divide Q into P until a remainder R(x) is obtained
such that
deg(R) < deg(Q)
Then
P (x) R(x)
f (x) = = S(x) +
Q(x) Q(x)
where S and R are also polynomials.

Example
ˆ
x3 + x
Find dx.
x−1
Solution.
ˆ 3 ˆ  
x +x 2
dx = x2 + x + 2 + dx
x−1 x−1
x3 x2
= + + 2x + 2 ln |x − 1| + C 1
3 2
2nd step
Factor the denominator Q(x) as far as possible.

3rd step
R(x)
The third step is to express the proper rational function as a sum of
Q(x)
partial fractions.
A theorem in algebra guarantees that it is always possible to do this. We
explain the details for the four cases that occur.

2
CASE 1: The denominator Q(x) is a product of distinct linear factors.
This means that we can write

Q(x) = (a1 x + b1 )(a2 x + b2 ) · · · (ak x + bk )

where no factor is repeated (and no factor is a constant multiple of another). In

this case, the partial fraction theorem states that there exist constants
A1 , A2 , · · · , Ak such that
R(x) A1 A2 Ak
= + +···+ . (1)
Q(x) a1 x + b1 a2 x + b2 ak x + bk

3
Example ˆ
x2 + 2x − 1
Evaluate dx.
2x3 + 3x2 − 2x
Solution.

2x3 + 3x2 − 2x = x(2x2 + 3x − 2) = x(2x − 1)(x + 2)

x2 + 2x − 1 A B C
= + +
x(2x − 1)(x + 2) x 2x − 1 x+2

x2 + 2x − 1 = A(2x − 1)(x + 2) + Bx(x + 2) + Cx(2x − 1)

= (2A + B + 2C)x2 + (3A + 2B − C)x − 2A

2A + B + 2C = 1, 3A + 2B − C = 2, −2A = −1
Thus
1 1 1
A= , B= , C=−
2 5 10
ˆ ˆ 
x2 + 2x − 1

1 1 1 1 1 1
dx = · + · − · dx
2x3 + 3x2 − 2x 2 x 5 2x − 1 10 x + 2
1 1 1
= · ln |x| + · ln |2x − 1| − · ln |x + 2| + K
2 10 10
4
Example
ˆ
dx
Find , where a ̸= 0.
x2 − a2
Solution.
1 1 A B
− = +
x2 − a2 (x − a)(x + a) x−a x+a
A(x + a) + B(x − a) = 1
Put x = a.
1
2a · A = 1 ⇒ A=
2a
Put x = −a.
1
2a · B = 1 ⇒ B=−
2a
ˆ ˆ  
dx 1 1 1
= − dx
x2 − a2 2a x−a x+a
1
= (ln |x − a| − ln |x + a|) + C
2a
1 x−a
= ln +C
2a x+a

5
CASE 2: Q(x) is a product of linear factors, some of which are repeated.
Suppose the first linear factor (a1 x + b1 ) is repeated r times; that is,
(a1 x + b1 )r occurs in the factorization of Q(x). Then instead of the single term
A1
in Equation (1), we would use:
(a1 x + b1 )

A1 A2 Ar
+ 2 +···+ r (2)
a1 x + b1 (a1 x + b1 ) (a1 + b1 )
x

For instance,
x3 − x + 1 A B C D E
3 = + 2 + + 2 +
x (x − 1)
2 x x x − 1 (x − 1) (x − 1)3

6
Example
ˆ
x4 − 2x2 + 4x + 1
Find dx.
x3 − x2 − x + 1
Solution.
1st step

x4 − 2x2 + 4x + 1 4x
=x+1+ 3
x3 − x2 − x + 1 x − x2 − x + 1
2nd step

x3 − x2 − x + 1 = (x − 1)(x2 − 1) = (x − 1)2 (x + 1)
3rd step

4x A B C
= + +
(x − 1)2 (x + 1) x−1 (x − 1)2 x+1

4x = A(x − 1)(x + 1) + B(x + 1) + C(x − 1)2

= (A + C)x2 + (B − 2C)x + (−A + B + C)
A + C = 0
B − 2C = 4
−A + B + C = 0
Thus
A = 1, B = 2, C = −1
7
4th step
ˆ 4 ˆ 
x − 2x2 + 4x + 1

1 2 1
dx = x + 1 + + − dx
x3 − x2 − x + 1 x−1 (x − 1)2 x+1
x2 2
= + x + ln |x − 1| − − ln |x + 1| + K
2 x−1
x2 2 x−1
= +x− + ln +K
2 x−1 x+1

8
CASE 3: Q(x) contains irreducible quadratic factors, none of which is
repeated.
If Q(x) has the factor ax2 + bx + c, where b2 − 4ac < 0, then, in addition to
R(x)
the partial fractions in Equations (1) and (2), the expression for will have
Q(x)
a term of the form
Ax + B
(3)
ax2 + bx + c
where A and B are constants to be determined.
For instance,
x A Bx + C Dx + E
= + 2 + 2
(x − 2)(x2 + 1)(x2 + 4) x−2 x +1 x +4

9
Example ˆ
2x2 − x + 4
Evaluate dx.
x3 + 4x
Solution. 2x2 − x + 4 A Bx + C
= + 2
x(x2 + 4) x x +4
2x2 − x + 4 = A(x2 + 4) + (Bx + C)x
= (A + B)x2 + Cx + 4A
A + B = 2, C = −1, 4A = 4 ⇒ A = 1, B = 1, C = −1
ˆ ˆ 
2x2 − x + 4

1 x−1
dx = + dx
x3 + 4x x x2 + 4
ˆ ˆ ˆ
x−1 x 1
dx = dx − dx
x2 + 4 x2 + 4 x2 + 4
2
Let u = x + 4. Then du = 2x dx.
ˆ ˆ
x 1 1 1 1
dx = · du = ln |u| + K1 = ln(x2 + 4) + K1
x2 + 4 2 u 2 2
ˆ  
1 1 −1 1
dx = tan x + K2
x2 + 4 2 2
ˆ ˆ ˆ ˆ
2x2 − x + 4 1 x 1
2
dx = dx + 2
dx − dx
x(x + 4) x x +4 x2 + 4
1 1
= ln |x| + ln(x2 + 4) − tan−1 (x/2) + K 10
2 2
Example ˆ
4x2 − 3x + 2
Evaluate dx.
4x2 − 4x + 3
Solution.
4x2 − 3x + 2 x−1
=1+ 2 (b2 − 4ac = −32 < 0)
4x2 − 4x + 3 4x − 4x + 3
x−1
=1+
(2x − 1)2 + 2
Let u = 2x − 1. Then du = 2 dx.
ˆ ˆ 
4x2 − 3x + 2

x−1
dx = 1 + dx
4x2 − 4x + 3 4x2 − 4x + 3
ˆ 1 ˆ
1 2
(u + 1) − 1 1 u−1
=x+ du = x + du
2 u2 + 2 4 u2 + 2
ˆ ˆ
1 u 1 1
=x+ du − du
4 u2 + 2 4 u2 + 2
 
1 1 1 u
= x + ln(u2 + 2) − · √ tan−1 √ +C
8 4 2 2
 
1 1 2x − 1
= x + ln(4x2 − 4x + 3) − √ tan−1 √ +C
8 4 2 2
11
Above example illustrates the general procedure for integrating a partial
fraction of the form
Ax + B
where b2 − 4ac < 0.
ax2 + bx + c

We complete the square in the denominator and then make a substitution that
brings the integral into the form
ˆ ˆ ˆ
Cu + D u 1
du = C du + D du
u2 + a2 u2 + a2 u2 + a2
Then the first integral is a logarithm and the second is expressed in terms of
tan−1 .

12
CASE 4: Q(x) contains a repeated irreducible quadratic factor.
Suppose Q(x) has the factor (ax2 + bx + c)r , where b2 − 4ac < 0, then instead
of the single partial fraction (3), the sum
A1 x + B1 A2 x + B2 Ar x + Br
+ +···+ (4)
ax2 + bx + c (ax2 + bx + c)2 (ax2 + bx + c)r

R(x)
occurs in the partial fraction decomposition of . Each of the terms in (4)
Q(x)
can be integrated by first completing the square.
For instance,
x3 + x2 + 1
x(x − 1)(x2 + x + 1)(x2 + 1)3
A B Cx + D Ex + F Gx + H Ix + J
= + + 2 + 2 + +
x x−1 x +x+1 x +1 (x2 + 1)2 (x2 + 1)3

13
Example ˆ
1 − x + 2x2 − x3
Evaluate dx.
x(x2 + 1)2
Solution.
1st step
2nd step
3rd step 1 − x + 2x2 − x3 A Bx + C Dx + E
= + 2 +
x(x2 + 1)2 x x +1 (x2 + 1)2
−x3 + 2x2 − x + 1 = A(x2 + 1)2 + (Bx + C)x(x2 + 1) + (Dx + E)x
= A(x4 + 2x2 + 1) + B(x4 + x2 ) + C(x3 + x) + Dx2 + Ex
= (A + B)x4 + Cx3 + (2A + B + D)x2 + (C + E)x + A
A + B = 0, C = −1, 2A + B + D = 2, C + E = −1, A = 1
Thus
A = 1, B = −1, C = −1, D = 1, E = 0
ˆ 2 3 ˆ  
1 − x + 2x − x 1 x+1 x
dx = − + dx
x(x2 + 1)2 x x2 + 1 (x2 + 1)2
ˆ ˆ ˆ ˆ
dx x dx x dx
= − dx − +
x x2 + 1 x2 + 1 (x + 1)2
2

1 1
= ln |x| − ln(x2 + 1) − tan−1 x − +K
2 2(x2 + 1) 14
Example
ˆ
x2 + 1
Find dx.
x(x2 + 3)
Solution. Let u = x3 + 3x. Then du = (3x2 + 3) dx = 3(x2 + 1) dx.
ˆ ˆ
x2 + 1 1 1 1 1
dx = dx = ln |u| + C = ln |x3 + 3x| + C
x(x2 + 3) 3 u 3 3

15
Remark
ˆ
1 1
(1) dx = ln |ax + b| + C
ax + b a
ˆ
1 1
(2) dx = (ax + b)1−m + C, (m > 1)
(ax + b)m a(1 − m)

Remark

b

e b
dx + e d x + 2a + d − 2a
2 m
= m
 m
(ax + bx + c) a b 2
(x + 2a ) + 4ac−b
2
4a2

u+ℓ
:= A
(u2 + k2 )m
A 2u 1
= + Aℓ 2
2 (u2 + k2 )m (u + k2 )m

16
Remark
ˆ
2u
(1) du = ln |u2 + k2 | + C
u2 + k 2
ˆ
2u 1
(2) du = (u2 + k2 )1−m + C, (m > 1)
(u2 + k2 )m (1 − m)
ˆ
1 1 u
(3) 2 2
du = tan−1 +C
u +k k k
(4)
ˆ
1
du
(u2 + k2 )m
ˆ
1 u 2m − 3 1
= + du (m > 1)
2(m − 1)k2 (u2 + k2 )m−1 2(m − 1)k2 (u2 + k2 )m−1

Proof.
ˆ
1
du
(u2 + k2 )m−1
ˆ
1 u2
=u 2 + 2(m − 1) du
(u + k2 )m−1 (u2 + k 2 )m
ˆ ˆ
1 u2 + k 2 1
=u + 2(m − 1) du − 2(m − 1)k2 du
(u2 + k2 )m−1 (u2 + k2 )m (u2 + k2 )m
17
Rationalizing substitutions
Some nonrational functions can be changed into rational functions by means of
appropriate substitutions.
p In particular, when an integrand p
contains an
expression of the form n g(x), then the substitution u = n g(x) may be
effective.

Example ˆ √
x+4
Evaluate dx.
x
√
Solution. Let u = x + 4. Then u2 = x + 4, so x = u2 − 4 and dx = 2u du
ˆ √ ˆ ˆ
x+4 u u2
dx = 2u du = 2 du
x u2 − 4 u2 − 4
ˆ  
4
=2 1+ 2 du
u −4
ˆ ˆ
du
= 2 du + 8
u2 − 4
1 u−2
= 2u + 8 · ln +C
2·2 u+2
√
√ x+4−2
= 2 x + 4 + 2 ln √ +C
x+4+2
18