Integrating Rational Functions by Partial Fractions
P( x)
is a rational function; that is, P (x ) and Q (x) are polynomial
Q( x )
functions. If the degree of P (x ) is greater than or equal to the degree of Q (x) , then
R( x)
P( x)
R( x)
by long division, f ( x) 
where
is a proper rational fraction;
 S ( x) 
Q ( x)
Q( x)
Q( x)
that is, the degree of R (x ) is less than the degree of Q (x) . A theorem in advanced
algebra states that every proper rational function can be expressed as a sum
Suppose f ( x) 
R ( x)
 F1 ( x)  F2 ( x)      Fn ( x)
Q( x)
where F1 ( x), F2 ( x), ..., Fn ( x) are rational functions of the form
Ax  B
A
or
2
k
(ax  bx  c) k
(ax  b)
in which the denominators are factors of Q (x) . The sum is called the partial fraction
R( x)
decomposition of
. The first step is finding the form of the partial fraction
Q ( x)
R( x)
decomposition of
is to factor Q (x) completely into linear and irreducible quadratic
Q ( x)
factors, and then collect all repeated factors so that Q (x) is expressed as a product of
distinct factors of the form
(ax  b) m and (ax 2  bx  c) m .
From these factors we can determine the form of the partial fraction decomposition using
the following two rules:
Linear Factor Rule: For each factor of the form (ax  b) m , the partial fraction
decomposition contains the following sum of m partial fractions:
Am
A1
A2

  
2
ax  b (ax  b)
(ax  b) m
where A1, A2, . . ., Am are constants to be determined.
1
Quadratic Factor Rule: For each factor of the form (ax 2  bx  c) m , the partial
fraction decomposition contains the following sum of m partial fractions:
Am x  Bm
A1 x  B1
A2 x  B2

  
2
2
2
ax  bx  c (ax  bx  c)
(ax 2  bx  c) m
where A1, A2, . . ., Am, B1, B2, …, Bm are constants to be determined.
I. Integrating Proper Rational Functions
Example 1: Find

3x  17
dx .
x  2x  3
2
x 2  2 x  3  ( x  3)( x  1)  using the Linear Factor Rule, we get
3x  17
3x  17
A
B



 3x  17  A( x  1)  B( x  3) after
x  2 x  3 ( x  3)( x  1) x  3 x  1
2
multiplying by ( x  3)( x  1) . If we let x  3 , then  8  4 A  A  2 ; if we let
x  1 , then  20  4B  B  5 . Thus,

3x  17
dx  2
( x  3)( x  1)
Example 2: Find


1
dx  5
x 3


3x  17
dx =
x  2x  3
2
1
dx   2 ln x  3  5 ln x  1  C .
x 1
3x  4
dx .
x  4x  4
2
x 2  4 x  4  ( x  2)( x  2)  ( x  2) 2  by the Linear Factor Rule, we get
B
3x  4
3x  4
A
 3 x  4  A( x  2)  B after multiplying



2
x  2 ( x  2) 2
x  4 x  4 ( x  2)
2
by ( x  2) 2 . If we let x = 2, then 2  B ; if we let x = 3, then 5  A  B  A  2 
A  3 . Thus,
3 ln x  2 


3x  4
dx  3
x  4x  4
2
1
dx  2
x2
2
C.
x2
2

1
dx 
( x  2) 2
Example 3: Find

4x2  x  2
dx .
x3  x2
x 3  x 2  x 2 ( x  1)  using the Linear Factor Rule with both x2 (multiplicity of
4x 2  x  2 4x 2  x  2 A B
C
linear factors) and x – 1, we get
 2
  2

3
2
x x
x 1
x x
x ( x  1)
4 x 2  x  2  Ax( x  1)  B( x  1)  Cx 2 after multiplying by x 2 x  1 . If we let
x = 0, then  2  B(1)  B  2 ; if we let x = 1, then C  3 ; and, if we let x = 2,
then 16  2 A  B  4C  2 A  2  12  2  2 A  A  1. Thus,

4x 2  x  2
dx 
x3  x2
Example 4: Find

x 1
dx 
x2 1

1
dx  2
x

x 1
dx .
x2 1

x
dx 
x 1
2



1
dx  3
x2
1
1
dx 
2
x 1
2
1
2
dx  ln x   3 ln x 1  C .
x 1
x

2x
dx 
x 1
2

1
dx 
x 1
2
1
ln x 2  1  arctan x  C . [Note: This problem doesn’t really illustrate the
2
Quadratic Factor Rule, but it does illustrate how to split a fraction with a linear
numerator and a quadratic denominator into the sum or difference of two fractions
with the same quadratic denominator.]
Example 5: Find

7x2  x  2
dx .
( x  1)( x 2  1)
Using both the Linear Factor Rule and the Quadratic Factor Rule, we get
7x2  x  2
A
Bx  C

 2
 7 x 2  x  2  A( x 2  1)  ( Bx  C )( x  1)
2
( x  1)( x  1) x  1 x  1
after multiplying by ( x  1)( x 2  1) . If we let x  1 , then 10  2 A  A  5 ;
3
if we let x  0 , then 2  A  C  5  C  C  3; and if we let x  2 , then 32 =
5 A  2B  C  25  2B  3  2B  4  B  2 . Thus,

5
dx 
x 1

2x  3
dx  5
x2 1

1
dx 
x 1



2x
3
x 1
2
7x2  x  2
dx =
( x  1)( x 2  1)
1
dx 
x 1
2
5 ln x  1  ln x 2  1  3 arctan x  C .
Example 6: Find

5 x 2  11
dx .
( x 2  1)( x 2  4)
5 x 2  11
Ax  B Cx  D
 2

 5 x 2  11  ( Ax  B)( x 2  4) 
2
2
( x  1)( x  4) x  1 x 2  4
(Cx  D)( x 2  1) after multiplying by ( x 2  1)( x 2  4) . If we let x = 0, then
11  4 B  D ; if we let x  1 , then 16  5 A  5B  2C  2D ; if we let x  1 ,
then 16  5 A  5B  2C  2D ; and if x  2 , then 31  16 A  8B  10C  5D .
If we add the equations 16  5 A  5B  2C  2D and 16  5 A  5B  2C  2D ,
we get 32  10B  4D  combining this equation with 11  4 B  D , we get
B  2 and D  3 . If we subtract the equations 16  5 A  5B  2C  2D and
16  5 A  5B  2C  2D , we get 0  10 A  4C  2C  5 A  combining this
equation with 31  16 A  8B  10C  5D , B  2, and D  3 , we get 31 = 16A +
16  25 A  15  A  0  C  0 . Thus,

3

5 x 2  11
dx  2
( x 2  1)( x 2  4)

1
dx 
x 1
2
1
3
 x
dx  2 arctan x  arctan    C .
2
x 4
2
2
Example 7: Find

1
dx .
x  x2
4
1
1
A B Cx  D
 2 2
  2 2
 1  Ax( x 2  1)  B( x 2  1) 
2
x
x x
x ( x  1)
x
x 1
4
(Cx  D) x 2 after multiplying by x 2 ( x 2  1) . If we let x  0 , then 1  B. If we
let x  1 , then 1  2 A  2B  C  D  2 A  C  D  1. If we let x  1 , then
1  2 A  2B  C  D  2 A  C  D  1  adding this equation to the equation
4
2 A  C  D  1 , we get D  1 . If we let x  2 , then 1  10 A  5B  8C  4D 
0  10 A  8C . Taking the equations 0  10 A  8C and 2 A  C  0 , we obviously
1
1
1
get that A  0 and C  0 . Thus,
dx 
dx 
dx 
4
2
2
2
x x
x
x 1
1
 arctan x  C .
x



II. Integrating Improper Rational Functions
Although the method of partial fractions only applies to proper rational functions,
an improper rational function can be integrated by performing long division (or
P( x)
synthetic division). If f ( x) 
is a rational function where P (x ) and Q (x) are
Q( x )
polynomial functions and the degree of P (x ) is greater than or equal to the degree of
R( x)
P( x)
R( x)
Q (x ) , then by long division, f ( x) 
where
is a proper
 S ( x) 
Q ( x)
Q( x)
Q( x)
R( x)
rational fraction. Since
is a proper rational function, it can be decomposed
Q ( x)
into partial fractions.
Example 1: Find

x3 1
dx .
x 1
By synthetic division, 1

x3 1
dx 
Thus,
x 1
ln x 1  C .
Example 2: Find


1
x3 1
2
 x2  x 1
0 0 1 
.
x 1
x 1
1 1 1
1
1
1
2
( x 2  x  1) dx  2

1
1
1
dx  x 3  x 2  x 
x 1
3
2
x3
dx .
x2  x  2
x3
3x  2
 x 1 2
By long division (see next page), 2
. Thus,
x x2
x x2
x3
3x  2
dx  ( x  1) dx 
dx . By partial fraction
2
2
x x2
x  x2



5
decomposition,

8
1
3x  2
dx  ln x  2  ln x  1 
2
3
3
x  x2

x3
dx =
x2  x  2
1 2
8
1
x  x  ln x  2  ln x  1  C .
2
3
3
x + 1
x2  x  2
x3  0x 2  0x  0
x3  x 2  2x
x 2  2x
x2  x  2
3x  2
Practice Sheet for Rational Functions Using Partial
Fraction Decomposition
(1)

 x 2  3x  4
(2)

4x  2
(3)

(4)

(5)

x ( x  2) 2
( x  1)( x 2  1)
x6
2
x  2x
1
u2  a2
dx =
dx =
dx =
du =
3x 2  x  1
( x  1)( x 2  4)
dx =
6
(6)

(7)

(8)

(9)

(10)

6x 2  7x  6
(11)

ex
(12)

(13)

(14)

(15)

(16)

(17)

3x 2  x  2
( x  1)( x 2  1)
ex
e 2 x  3e x  2
1
x x 1
x3  1
2
x 1
dx =
dx =
dx =
dx =
x 2 ( x  2)
e2x  e x  6
dx =
dx 
sin x
cos 2 x  3 cos x  2
dx 
5 x 2  4 x  12
( x  1)( x 2  2 x  10)
e2x
e2x  e x  6
e3x
e2x  e x  6
dx 
dx 
dx 
x2 1
dx 
x
1
2
x x 4
dx 
7
Solution Key for Rational Functions Using Partial Fraction
Decomposition
(1)
 x 2  3x  4 A
B
C
 

  x 2  3x  4  A( x  2) 2  Bx ( x  2) 
2
2
x x  2 ( x  2)
x( x  2)
Cx after multiplying by x( x  2) 2 . If we let x  0 , then 4  4 A  A  1 ; if we let
x  2 , then  6  2C  C  3; and if we let x  1 , then 0  A  B  C  B 
 x 2  3x  4
1
1
dx 
dx  2
dx 
A  C  1  3  2 . Thus,
2
x
x2
x( x  2)

3
(2)



1
3
dx  ln x  2 ln x  2 
C.
2
x2
( x  2)
4x  2
A
Bx  C

 2
 4 x  2  A( x 2  1)  ( Bx  C )( x  1) after
2
( x  1)( x  1) x  1 x  1
multiplying by ( x  1)( x 2  1) . If we let x  1, then 6  2 A  A  3 ; if we let
x  0 , then 2  A  C  C  1 ; and if we let x  2 , then 10  5 A  2B  C 
4x  2
1
dx  3
dx 
10  15  2B  1  B  3 . Thus,
2
x 1
( x  1)( x  1)



 3x  1
1
3
dx  3
dx 
2
x 1
2
x 1


2x
dx 
x 1
2

1
dx  3 ln x  1 
x 1
2
3
ln x 2  1  arctan x  C .
2
(3)
x6
x6
A
B

 
 x  6  A( x  2)  Bx after multiplying by
2
x  2 x x( x  2) x x  2
x( x  2) . If we let x  0 , then  6  2 A  A  3 ; and if we let x  2 , then
x6
1
1
dx  3
dx  2
dx  3 ln x 
 4  2B  B  2 . Thus,
2
x
x2
x  2x
2 ln x  2  C .

(4)


1
1
A
B



 1  A(u  a)  B(u  a) after
2
(u  a)(u  a) u  a u  a
u a
1
multiplying by (u  a )(u  a ) . If we let u  a , then 1  A(2a )  A 
; and
2a
2
8

1
1
. Thus,
du 
2
2a
u  a2
1
1
1
du 
ln u  a  ln u  a  C 
ua
2a
2a
if we let u  a , then 1  B(2a )  B  

1
1
1
du 
2a u  a
2a
1
ua
ln
C.
2a u  a
(5)

3x 2  x  1
A
Bx  C

 2
 3x 2  x  1  A( x 2  4)  ( Bx  C )( x  1) after
2
( x  1)( x  4) x  1 x  4
multiplying by ( x  1)( x 2  4). If we let x  1, then 5  5 A  A  1 ; if we let
x  0 , then 1  4 A  C  C  4(1)  1  3 ; and if we let x  2 , then 15  8 A  2B 

(6)
1
dx 
x 1


3x 2  x  2
dx 
( x  1)( x 2  1)


1
2x  3
dx 
dx 
x 1
x2  4
2x
1
3
 x
dx  3 2
dx  ln x  1  ln x 2  4  arctan    C .
2
2
x 4
x 4
2
C  8  2B  3  B  2 . Thus,

3x 2  x  2
A
Bx  C

 2
 3x 2  x  2  A( x 2  1)  ( Bx  C )( x  1) after
2
( x  1)( x  1) x  1 x  1
multiplying by ( x  1)( x 2  1) . If we let x  1, then 2  2 A  A  1 ; if we let
x  0 , then  2  A  C  C  A  2  1  2  3 ; and if we let x  2 , then
3x 2  x  2
1
dx 
dx 
12  5 A  2B  C  5  2B  3  B  2 . Thus,
2
x 1
( x  1)( x  1)


2x  3
dx 
x2 1

1
dx 
x 1



2x
1
dx  3 2
dx  ln x  1  ln x 2  1 +
x 1
x 1
2
3 arctan x  C .
(7)


ex
1
dx 
du 
2x
x
2
e  3e  2
u  3u  2
1
1
A
B
du.


 1  A(u  2)  B(u  1) after
(u  1)(u  2)
(u  1)(u  2) u  1 u  2
multiplying by (u  1)(u  2) . If we let u  1 , then 1  A; and if we let u  2 ,
1
1
1
then 1  B(1)  B  1. Thus,
du 
du 
du 
(u  1)(u  2)
u 1
u2
Let u  e x  du  e x dx 


ln u  1  ln u  2  C  ln

u 1
e 1
 C  ln x
 C.
u2
e 2
x
9

 x x  1 dx 
1
2udu
1
2
du . Using the result in problem 4, 2
 (u  1) u  u  1
 u 1 du =
1
(8) Let u  x  1  u 2  x  1  x  u 2  1  dx  2udu 
2
2
2
1 u 1 
2 ln
  C  ln
2 u 1 
x 1 1
x 1 1
C .

x3  1
 x 1
x3  1

x


dx 
x2 1
x2 1
x2 1
1
1
1
dx  x 2  ln x 2  1  arctan x  C .
2
2
2
x 1
(9) Use long division of polynomials to get

(10)
x dx 
1
2

2x
dx 
x 1
2

6x 2  7 x  6 A B
C
  2 
 6 x 2  7 x  6  Ax( x  2)  B( x  2)  Cx 2 after
2
x x
x2
x ( x  2)
multiplying by x 2 ( x  2) . If we let x  0 , then 6  B(2)  B  3 ; if we let
x  2 , then 16  4C  C  4 ; and if we let x  3 , then 39  3A  B  9C 
6x 2  7x  6
1
1
dx  2
dx  3 2 dx 
39  3A  3  36  A  2 . Thus,
2
x
x ( x  2)
x
1
3
4
dx  2 ln x   4 ln x  2  C .
x2
x


(11) Let u  e x  du  e x dx 
e
2x

ex
dx 
 ex  6
u
2

1
du 
u 6

1
1
A
B
du .


 1  A(u  2)  B(u  3) after
(u  3)(u  2)
(u  3)(u  2) u  3 u  2
1
multiplying by (u  3)(u  2) . If we let u  3 , then 1  5 A  A  ; and if we let
5
x
e
1
1
dx 
du 
u  2 , then 1  5 B  B   . Thus,
2x
x
2
5
e e 6
u u 6

1
5

1
1
du 
u 3
5


1
1
1
1 ex  3
du  ln u  3  ln u  2  C  ln x
C.
u2
5
5
5 e 2
(12) Let u  cos x  du   sin x dx 


sin x
1
dx  
du .
(u  1)(u  2)
cos x  3 cos x  2
2
1
A
B


 1  A(u  2)  B(u  1) after multiplying by
(u  1)(u  2) u  1 u  2
10
(u  1)(u  2) . If we let u  2 , then 1  B ; and if we let u  1, then 1   A 
sin x
1
1
A  1 . Thus,
dx  
du 
du 
2
(u  1)(u  2)
u 1
cos x  3 cos x  2
1
u 1
cos x  1
du  ln u  1  ln u  2  C  ln
 C  ln
C .
u2
u2
cos x  2


(13)


5 x 2  4 x  12
A
Bx  C

 2
 5 x 2  4 x  12  A( x 2  2 x  10) 
2
( x  1)( x  2 x  10) x  1 x  2 x  10
( Bx  C )( x  1) after multiplying by ( x  1)( x 2  2 x  10) . If we let x  1, then
13  13A  A  1 ; if we let x  0 , then 12  10 A  C  C  2 ; and if we let
x  2 , then 24  18 A  2B  C  18  2B  2  B  4 . Thus,
5 x 2  4 x  12
1
4x  2
1
dx 
dx 
dx 
dx 
2
2
x 1
x 1
( x  1)( x  2 x  10)
x  2 x  10
4x  4  6
1
4x  4
1
dx 
dx 
dx  6 2
dx 
2
2
x 1
x  2 x  10
x  2 x  10
x  2 x  10
1
2x  2
1
dx  2
dx  6
dx  ln x  1  2 ln x 2  2 x  10 
2
x 1
x  2 x  10
( x  1) 2  9











 x 1
2 arctan 
C.
 3 



e2x
ex
dx

e x dx 
e2x  e x  6
e2x  e x  6
u
u
A
B
du .


 u  A(u  2)  B(u  3) after
(u  3)(u  2)
(u  3)(u  2) u  3 u  2
(14) Let u 2  e 2 x  u  e x  du  e x dx 
3
; and if we let
5
e2x
ex
2
dx

e x dx 
u  2 , then  2  5 B  B  . Thus,
2x
x
2x
x
5
e e 6
e e 6
u
3
1
2
1
3
2
du 
du 
du  ln u  3  ln u  2  C 
(u  3)(u  2)
5 u 3
5 u2
5
5
3
2
ln e x  3  ln e x  2  C .
5
5
multiplying by (u  3)(u  2) . If we let u  3 , then 3  5 A  A 




(15) Let u 2  e 2 x  u  e x  du  e x dx 
e
11
2x

e3x
dx 
 ex  6
e
2x
e2x
e x dx 
 ex  6




u2
u6 
u6

du  1  2
du.
 du  1 du 
2
(u  3)(u  2)
u u 6
 u u 6
u6
A
B


 u  6  A(u  2)  B(u  3) after multiplying by
(u  3)(u  2) u  3 u  2
9
(u  3)(u  2) . If we let u  3 , then 9  5 A  A  ; and if we let u  2 , then
5
3x
e
9
1
4
dx  1 du 
du 
4  5 B  B   . Thus,
2x
x
5 u 3
5
e e 6
4
1
9
4
9
4
du  u  ln u  3  ln u  2  C  e x  ln e x  3  ln e x  2  C
5 u2
5
5
5
5




.
(16) Let u  x 2  1  u 2  x 2  1  2udu  2 xdx  udu  xdx 


 u 1 
1
1 u 1
1
x
 C  x  1  ln
du  u  ln
 u 1
2 u 1
2
x
x 1
x dx 
x2
2
u
udu  
u 1
2
u
2
du 
2

1 

1  2
 du  1 du 
 u 1
2
1 1
2
1 1
2
2

C.
(17) Let u  x 2  4  u 2  x 2  4  2u du  2 x dx  u du  x dx 
x
1
2
1
ln
4
x 4
2
x dx 
x2  4  2
x2  4  2

1
u du 
(u  4)(u )
2
C.
12

x2 1
dx 
x
x
1
x2  4
1
1 u2
du  ln
C 
4 u2
u 4
2
dx =