Higher Engineering Mathematics

INTEGRATION USING PARTIAL FRACTIONS 411

H

It was shown in Problem 9, page 23:

3 + 6 x+ 4 x^2 − 2 x^3

x^2 (x^2 +3)

≡

2

x

+

1

x^2

+

3 − 4 x

(x^2 +3)

Thus

∫

3 + 6 x+ 4 x^2 − 2 x^3

x^2 (x^2 +3)

dx

≡

∫ (

2

x

+

1

x^2

+

(3− 4 x)

(x^2 +3)

)

dx

=

∫ {

2

x

+

1

x^2

+

3

(x^2 +3)

−

4 x

(x^2 +3)

}

dx

∫
3
(x^2 +3)

dx= 3

∫

1

x^2 +(

√

3)^2

dx

=

3

√

3

tan−^1

x

√

3

, from 12, Table 40.1, page 398.

∫
4 x
x^2 + 3

dxis determined using the algebraic sub-

stitutionu=(x^2 +3).

Hence

∫ {

2

x

+

1

x^2

+

3

(x^2 +3)

−

4 x

(x^2 +3)

}

dx

=2lnx−

1

x

+

3

√

3

tan−^1

x

√

3

−2ln(x^2 +3)+c

=ln

(

x

x^2 + 3

) 2

−

1

x

+

√

3 tan−^1

x

√

3

+c

Problem 9. Determine

∫

1

(x^2 −a^2 )

dx.

Let

1

(x^2 −a^2 )

≡

A

(x−a)

+

B

(x+a)

≡

A(x+a)+B(x−a)

(x+a)(x−a)

Equating the numerators gives:

1 ≡A(x+a)+B(x−a)

Let x=a, then A=

1

2 a

, and let x=−a, then

B=−

1

2 a

Hence

∫

1

(x^2 −a^2 )

dx

≡

∫

1

2 a

[

1

(x−a)

−

1

(x+a)

]

dx

=

1

2 a

[ln(x−a)−ln (x+a)]+c

=

1

2 a

ln

(

x−a

x+a

)

+c

Problem 10. Evaluate

∫ 4

3

3

(x^2 −4)

dx,

correct to 3 significant figures.

From Problem 9,

∫ 4

3

3

(x^2 −4)

dx= 3

[

1

2(2)

ln

(

x− 2

x+ 2

)] 4

3

=

3

4

[

ln

2

6

−ln

1

5

]

=

3

4

ln

5

3

= 0. 383 , correct to 3

significant figures

Problem 11. Determine

∫

1

(a^2 −x^2 )

dx.

Using partial fractions, let

1

(a^2 −x^2 )

≡

1

(a−x)(a+x)

≡

A

(a−x)

+

B

(a+x)

≡

A(a+x)+B(a−x)

(a−x)(a+x)

Then 1≡A(a+x)+B(a−x)

Letx=athenA=

1

2 a

. Letx=−athenB=

1

2 a

Hence

∫

1

(a^2 −x^2 )

dx

=

∫

1

2 a

[

1

(a−x)

+

1

(a+x)

]

dx