Higher Engineering Mathematics, Sixth Edition

Integration usingpartial fractions 411

7.

∫ 6

4

x^2 −x− 14

x^2 − 2 x− 3

dx [0.8122]

8. Determine the value of k, given that:
   ∫ 1

0

(x−k)

( 3 x+ 1 )(x+ 1 )

dx= 0

[

1

3

]

9. The velocity constantkof a given chemical
   reaction is given by:

kt=

∫ (

1

( 3 − 0. 4 x)( 2 − 0. 6 x)

)

dx

wherex=0whent=0. Show that:

kt=ln

{

2 ( 3 − 0. 4 x)

3 ( 2 − 0. 6 x)

}

41.3 Worked problemson

integration using partial

fractions with repeated linear

factors

Problem 5. Determine

∫

2 x+ 3

(x− 2 )^2

dx.

It was shown in Problem 5, page 16:

2 x+ 3

(x− 2 )^2

≡

2

(x− 2 )

+

7

(x− 2 )^2

Thus

∫

2 x+ 3

(x− 2 )^2

dx≡

∫ {

2

(x− 2 )

+

7

(x− 2 )^2

}

dx

=2ln(x−2)−

7

(x−2)

+c

⎡

⎣

∫

7

(x− 2 )^2

dxis determined using the algebraic

substitutionu=(x− 2 )— see Chapter 39.

⎤

⎦

Problem 6. Find

∫

5 x^2 − 2 x− 19

(x+ 3 )(x− 1 )^2

dx.

It was shown in Problem 6, page 16:

5 x^2 − 2 x− 19

(x+ 3 )(x− 1 )^2

≡

2

(x+ 3 )

+

3

(x− 1 )

−

4

(x− 1 )^2

Hence

∫

5 x^2 − 2 x− 19

(x+ 3 )(x− 1 )^2

dx

≡

∫ {

2

(x+ 3 )

+

3

(x− 1 )

−

4

(x− 1 )^2

}

dx

=2ln(x+3)+3ln(x−1)+

4

(x−1)

+c

orln

{

(x+3)^2 (x−1)^3

}

+

4

(x−1)

+c

Problem 7. Evaluate

∫ 1

− 2

3 x^2 + 16 x+ 15

(x+ 3 )^3

dx,

correct to 4 significant figures.

It was shown in Problem 7, page 17:

3 x^2 + 16 x+ 15

(x+ 3 )^3

≡

3

(x+ 3 )

−

2

(x+ 3 )^2

−

6

(x+ 3 )^3

Hence

∫

3 x^2 + 16 x+ 15

(x+ 3 )^3

dx

≡

∫ 1

− 2

{

3

(x+ 3 )

−

2

(x+ 3 )^2

−

6

(x+ 3 )^3

}

dx

=

[

3ln(x+ 3 )+

2

(x+ 3 )

+

3

(x+ 3 )^2

] 1

− 2

=

(

3ln4+

2

4

+

3

16

)

−

(

3ln1+

2

1

+

3

1

)

=− 0. 1536 ,correct to 4 significant figures.