Higher Engineering Mathematics, Sixth Edition

14 Higher Engineering Mathematics

The denominator factorizes as (x−1) (x+3) and the

numerator is of less degree than the denominator. Thus

11 − 3 x

x^2 + 2 x− 3

may be resolved into partial fractions.

Let

11 − 3 x

x^2 + 2 x− 3

≡

11 − 3 x

(x− 1 )(x+ 3 )

≡

A

(x− 1 )

+

B

(x+ 3 )

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 toeach

other.

Thus, 11− 3 x≡A(x+ 3 )+B(x− 1 )

To determine constantsAandB,valuesofxare 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.