Chapter 41
Integration using
partial fractions
41.1 Introduction
The process of expressing a fraction in terms of simpler
fractions—called partial fractions—is discussed in
Chapter 2, with the forms of partial fractions used being
summarized in Table 2.1, page 13.
Certain functions have to be resolved into partial fractions before they can be integrated as demonstrated in
the following worked problems.
41.2 Worked problems on
integration using partial
fractions with linear factors
Problem 1. Determine
11 −3x
x 2 + 2x − 3
dx.
As shown in Problem 1, page 13:
11 − 3x
x 2 + 2x − 3
≡
2
(x − 1)
−
5
(x + 3)
Hence
11 − 3x
x 2 + 2x − 3
dx
=
2
(x − 1)
−
5
(x + 3)
dx
= 2 ln(x −1)− 5 ln(x + 3)+ c
(by algebraic substitutions — see Chapter 39)
or ln
(x −1)
2
(x +3)
5
+ c by the laws of logarithms
Problem 2. Find
2x 2 − 9x − 35
(x + 1)(x − 2)(x + 3)
dx.
It was shown in Problem 2, page 14:
2x 2 − 9x − 35
(x + 1)(x − 2)(x + 3)
≡
4
(x + 1)
−
3
(x − 2)
+
1
(x + 3)
Hence
2x 2 − 9x − 35
(x + 1)(x − 2)(x + 3)
dx
≡
4
(x + 1)
−
3
(x − 2)
+
1
(x + 3)
dx
= 4 ln(x+ 1)− 3 ln(x− 2) + ln(x+ 3)+ c
or ln
(x + 1)
4 (x + 3)
(x −2)
3
+ c
Problem 3. Determine
x 2 + 1
x 2 − 3x + 2
dx.
Integration using
partial fractions
41.1 Introduction
The process of expressing a fraction in terms of simpler
fractions—called partial fractions—is discussed in
Chapter 2, with the forms of partial fractions used being
summarized in Table 2.1, page 13.
Certain functions have to be resolved into partial fractions before they can be integrated as demonstrated in
the following worked problems.
41.2 Worked problems on
integration using partial
fractions with linear factors
Problem 1. Determine
11 −3x
x 2 + 2x − 3
dx.
As shown in Problem 1, page 13:
11 − 3x
x 2 + 2x − 3
≡
2
(x − 1)
−
5
(x + 3)
Hence
11 − 3x
x 2 + 2x − 3
dx
=
2
(x − 1)
−
5
(x + 3)
dx
= 2 ln(x −1)− 5 ln(x + 3)+ c
(by algebraic substitutions — see Chapter 39)
or ln
(x −1)
2
(x +3)
5
+ c by the laws of logarithms
Problem 2. Find
2x 2 − 9x − 35
(x + 1)(x − 2)(x + 3)
dx.
It was shown in Problem 2, page 14:
2x 2 − 9x − 35
(x + 1)(x − 2)(x + 3)
≡
4
(x + 1)
−
3
(x − 2)
+
1
(x + 3)
Hence
2x 2 − 9x − 35
(x + 1)(x − 2)(x + 3)
dx
≡
4
(x + 1)
−
3
(x − 2)
+
1
(x + 3)
dx
= 4 ln(x+ 1)− 3 ln(x− 2) + ln(x+ 3)+ c
or ln
(x + 1)
4 (x + 3)
(x −2)
3
+ c
Problem 3. Determine
x 2 + 1
x 2 − 3x + 2
dx.
