Higher Engineering Mathematics, Sixth Edition

410 Higher Engineering Mathematics

By dividing out (since the numerator and denomina-

tor are of the same degree) and resolving into partial

fractions it was shown in Problem 3, page 14:

x^2 + 1

x^2 − 3 x+ 2

≡ 1 −

2

(x− 1 )

+

5

(x− 2 )

Hence

∫

x^2 + 1

x^2 − 3 x+ 2

dx

≡

∫{

1 −

2

(x− 1 )

+

5

(x− 2 )

}

dx

=(x−2) ln(x−1)+5ln(x−2)+c

orx+ln

{

(x−2)^5

(x−1)^2

}

+c

Problem 4. Evaluate

∫ 3

2

x^3 − 2 x^2 − 4 x− 4

x^2 +x− 2

dx,

correct to 4 significant figures.

By dividing out and resolving into partial fractions it

was shown in Problem 4, page 15:

x^3 − 2 x^2 − 4 x− 4

x^2 +x− 2

≡x− 3 +

4

(x+ 2 )

−

3

(x− 1 )

Hence

∫ 3

2

x^3 − 2 x^2 − 4 x− 4

x^2 +x− 2

dx

≡

∫ 3

2

{

x− 3 +

4

(x+ 2 )

−

3

(x− 1 )

}

dx

=

[

x^2

2

− 3 x+4ln(x+ 2 )−3ln(x− 1 )

] 3

2

=

(

9

2

− 9 +4ln5−3ln2

)

−( 2 − 6 +4ln4−3ln1)

=− 1. 687 ,correct to 4 significant figures.

Now try the following exercise

Exercise 162 Further problems on

integration using partial fractions with

linear factors

In Problems 1 to 5, integrate with respect tox.

1.

∫

12

(x^2 − 9 )

dx

⎡

⎢

⎣

2ln(x− 3 )−2ln(x+ 3 )+c

or ln

{

x− 3

x+ 3

} 2

+c

⎤

⎥

⎦

2.

∫

4 (x− 4 )

(x^2 − 2 x− 3 )

dx

⎡

⎢

⎢

⎣

5ln(x+ 1 )−ln(x− 3 )+c

or ln

{

(x+ 1 )^5

(x− 3 )

}

+c

⎤

⎥

⎥

⎦

3.

∫

3 ( 2 x^2 − 8 x− 1 )

(x+ 4 )(x+ 1 )( 2 x− 1 )

dx

⎡ ⎢ ⎢ ⎢ ⎢ ⎣

7ln(x+ 4 )−3ln(x+ 1 )

−ln( 2 x− 1 )+c or

ln

{

(x+ 4 )^7

(x+ 1 )^3 ( 2 x− 1 )

}

+c

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

4.

∫

x^2 + 9 x+ 8

x^2 +x− 6

dx

[

x+2ln(x+ 3 )+6ln(x− 2 )+c

orx+ln{(x+ 3 )^2 (x− 2 )^6 }+c

]

5.

∫

3 x^3 − 2 x^2 − 16 x+ 20

(x− 2 )(x+ 2 )

dx

⎡

⎣

3 x^2

2

− 2 x+ln(x− 2 )

−5ln(x+ 2 )+c

⎤

⎦

In Problems 6 and 7, evaluate the definite integrals

correct to 4 significant figures.

6.

∫ 4

3

x^2 − 3 x+ 6

x(x− 2 )(x− 1 )

dx [0.6275]