Higher Engineering Mathematics

PARTIAL FRACTIONS 19

A

whereAandBare constants to be determined,

i.e.

11 − 3 x

(x−1)(x+3)

≡

A(x+3)+B(x−1)

(x−1)(x+3)

,

by algebraic addition.
Since the denominators are the same on each side
of the identity then the numerators are equal to each
other.

Thus, 11− 3 x≡A(x+3)+B(x−1)

To determine constantsAandB, values ofxare
chosen to make the term inAorBequal to zero.

Whenx=1, then

11 −3(1)≡A(1+3)+B(0)

i.e. 8 = 4 A

i.e. A= 2

Whenx=−3, then

11 −3(−3)≡A(0)+B(− 3 −1)

i.e. 20 =− 4 B

i.e. B=− 5

Thus

11 − 3 x

x^2 + 2 x− 3

≡

2

(x−1)

+

− 5

(x+3)

≡

2

(x− 1 )

−

5

(x+ 3 )

[

Check:

2

(x−1)

−

5

(x+3)

=

2(x+3)−5(x−1)

(x−1)(x+3)

=

11 − 3 x

x^2 + 2 x− 3

]

Problem 2. Convert

2 x^2 − 9 x− 35

(x+1)(x−2)(x+3)

into the sum of three partial fractions.

Let

2 x^2 − 9 x− 35

(x+1)(x−2)(x+3)

≡

A

(x+1)

+

B

(x−2)

+

C

(x+3)

≡

(

A(x−2)(x+3)+B(x+1)(x+3)

+C(x+1)(x−2)

)

(x+1)(x−2)(x+3)

by algebraic addition.

Equating the numerators gives:

2 x^2 − 9 x− 35 ≡A(x−2)(x+3)

+B(x+1)(x+3)+C(x+1)(x−2)

Letx=−1. Then

2(−1)^2 −9(−1)− 35 ≡A(−3)(2)

+B(0)(2)+C(0)(−3)

i.e. − 24 =− 6 A

i.e. A=

− 24

− 6

= 4

Letx=2. Then

2(2)^2 −9(2)− 35 ≡A(0)(5)+B(3)(5)+C(3)(0)

i.e. − 45 = 15 B

i.e. B=

− 45

15

=− 3

Letx=−3. Then

2(−3)^2 −9(−3)− 35 ≡A(−5)(0)+B(−2)(0)

+C(−2)(−5)

i.e. 10 = 10 C

i.e. C= 1

Thus

2 x^2 − 9 x− 35

(x+1)(x−2)(x+3)

≡

4

(x+ 1 )

−

3

(x− 2 )

+

1

(x+ 3 )

Problem 3. Resolve

x^2 + 1

x^2 − 3 x+ 2

into partial

fractions.

The denominator is of the same degree as the

numerator. Thus dividing out gives:

1

x^2 − 3 x+ 2

)

x^2 + 1

x^2 − 3 x+ 2

—————

3 x− 1

———

For more on polynomial division, see Section 1.4,

page 6.