Higher Engineering Mathematics, Sixth Edition

(Nancy Kaufman) #1

412 Higher Engineering Mathematics


Now try the following exercise

Exercise 163 Further problems on
integration using partial fractions with
repeated linear factors
In Problems 1 and 2, integrate with respect
tox.

1.


4 x− 3
(x+ 1 )^2

dx
[
4ln(x+ 1 )+

7
(x+ 1 )

+c

]

2.


5 x^2 − 30 x+ 44
(x− 2 )^3

dx




5ln(x− 2 )+

10
(x− 2 )

2
(x− 2 )^2

+c





In Problems 3 and 4, evaluate the definite integrals
correct to 4 significant figures.

3.

∫ 2

1

x^2 + 7 x+ 3
x^2 (x+ 3 )

[1.663]

4.

∫ 7

6

18 + 21 x−x^2
(x− 5 )(x+ 2 )^2

dx [1.089]


  1. Show that


∫ 1

0

(
4 t^2 + 9 t+ 8
(t+ 2 )(t+ 1 )^2

)
dt= 2. 546 ,

correct to 4 significant figures.

41.4 Worked problemson


integration using partial


fractions with quadratic factors


Problem 8. Find


3 + 6 x+ 4 x^2 − 2 x^3
x^2 (x^2 + 3 )

dx.

It was shown in Problem 9, page 18:

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 399.


4 x
x^2 + 3

dxis determined using the algebraic substi-
tutionu=(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

+


3tan−^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,thenA=

1
2 a

,andletx=−a,then

B=−

1
2 a

Hence


1
(x^2 −a^2 )

dx



1
2 a

[
1
(x−a)

1
(x+a)

]
dx
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