MTH 252
Integral Calculus
Chapter 8 – Principles of
Integral Evaluation
Section 8.5 – Integrating Rational
Functions by
Partial Fractions
Copyright © 2006 by Ron Wallace, all rights reserved.
Review:
Addition/Subtraction of Fractions
a c ad  bc
 
b d
bd
Note: The equation works both ways!
Problem:
Find two fractions whose sum/difference
is equal to a third given fraction.
The product of the denominators of the
two fractions will be the denominator of
the given fraction.
Example:
ad  bc a c
 
bd
b d
Find two fractions whose
sum or difference is:
11
35
Denominator = 35
Possibilities for the other two denominators are:
1 & 35 and 5 & 7
11 A B
 
35 5 7
11  7 A  5B
Many solutions, including:
3 & -2 and 1 & 4/5
Rational Functions
Any function of the form,
P( x )
f ( x) 
Q( x)
where
P( x ) and Q ( x )
are polynomials.
Fundamental Theorem of Algebra


Every polynomial equation of degree n
with complex coefficients has n roots in
the complex numbers.

Each real root r, gives a factor (x-r).

Each complex root +i has a companion root
-i. These give a factor: (ax2+bx+c).
Hence, every polynomial can be written as
a product of linear & quadratic factors.
Some “Easy” Integrals
of Rational Functions

5
dx  5ln x  3  c
x3
5
 2 x  3 dx

Linear
Denominator
5 1
5
  du  ln 2 x  3  c
2 u
2
Let u=2x-3
du=2dx
4x
12
dx   4 
dx  4 x  12 ln x  3  c
x3
x3
17
4x  5
 x  3 dx   4  x  3 dx  4 x  17 ln x  3  c
5
 (2 x  3)7 dx
Let u=2x-3
du=2dx

5
2
1
 u 7 du 
5
2
7
6
6
5
5
u
du


u

c


(2
x

3)
c
12
12

Partial
Fractions
P( x )
 Q( x) dx
where Q(x) is a product
of linear factors &
deg(P(x)) < deg(Q(x))
2x  5
A
B


(3x  1)( x  4)
(3x  1) ( x  4)
2 x  5  A( x  4)  B(3x  1)  ( A  3B) x  (4 A  B)
2  A  3B
5  4 A  B
Solve this system
for A & B.
A=-1, B=1
2x  5
1
1
1
 (3x  1)( x  4)dx   3x  1  x  4 dx   3 ln 3x  1  ln x  4  c
This method can be extended to any
number of distinct linear factors.
Partial
Fractions
Repeated
Linear Factors
P( x )
 Q( x) dx
4x 1
(2 x  3) 2
where Q(x) is a product
of linear factors &
deg(P(x)) < deg(Q(x))
A
BB


(2 x  3) (2 x  3) 2
4 x  1  A(2 x  3)  B  2 Ax  (3 A  B)
4  2A
1  3 A  B
Solve this system
for A & B.
A=2, B=-7
4x 1
2
7
7
 (2 x  3)2 dx   2 x  3  (2 x  3)2 dx  ln 2 x  3  2(2 x  3)  c
This method can be extended to any power of the denominator
and can be combined with the previous method.
More “Easy” Integrals
of Rational Functions

Quadratic
Denominator
3
1
dx  3
dx
2
2
x  4x  7
( x  2)  3
1
 3 2
du
u 3
Let u=x-2
du=dx
Finish using trig substitutions.
3
2x  4
17
3 x  11
 x 2  4 x  7 dx  2  x 2  4 x  7 dx   x 2  4 x  7 dx
3
2
6  11
3x  11  _____(2
x  4)  _____
d 2
 x  4 x  7  
dx
Just like the
one above!
3
2
ln x 2  4 x  7
More “Easy” Integrals
of Rational Functions
Quadratic
Denominator
3
2x  4
17
3x  11

dx

dx
dx
2
5
2
5


2
5
 ( x  4 x  7)
2 ( x  4 x  7)
( x  4 x  7)
Complete the
square & trig
substitution.
Let u = x2 - 4x + 7
du = 2x – 4 dx
3 5
3 4
u
du


u c

2
8
3 2
  ( x  4 x  7) 4  c
8

Partial
P( x )
dx

Fractions Q( x)
4 x2  x  3
( x  1)( x 2  2 x  3)
where Q(x) is a product of a
linear factor & a quadratic
factor & deg(P(x)) < deg(Q(x))
A
BxB C

 2
( x  1) ( x  2 x  3)
4 x 2  x  3  ( A  B) x 2  (2 A  B  C ) x  (3 A  C )
4  A B
1  2 A  B  C
3  3A  C
Solve this system
for A, B, & C.
A=1, B=3, C=0
4 x 2  3x  3
1
3x
dx

dx

 ( x 1)( x2  2x  3)  x 1  x2  2x  3 dx
This method can be extended to any
number of distinct linear & quadratic factors.
Partial
Fractions
Repeated
Quadratic Factors
P( x )
 Q( x) dx
where Q(x) is a product
of quadratic factors &
deg(P(x)) < deg(Q(x))
P( x)
(3x 2  x  2)2
Ax  B
Cx  D


2
(3x  x  2) (3x 2  x  2)2
… and proceed as before!
All of these methods can be combined and extended to handle
any rational function where you can factor the denominator
into a product of linear and quadratic factors.
Partial
Fractions
Example …
where Q(x) is a product
of quadratic factors &
deg(P(x)) < deg(Q(x))
P( x )
 Q( x) dx
x  3x  2 x
 ( x4 1)( x3 1)( x3  1) dx
5
3
P( x )
 Q( x) dx
where deg(P(x)) ≥ deg(Q(x))
Simplify using long division of polynomials.
Example:
4 x 3  3x  5
 2 x  1 dx 
4
 2 x dx   xdx   dx   2 x  1 dx
2x  x  1
2
2

4
2x  1
2 x  1 4 x 3  0 x 2  3x  5
4 x3  2 x2
2 x 2  3x
2x 2  x
2 x  5
2 x  1
4