The Chemistry Maths Book, Second Edition

6.3 The method of substitution 169

(ii).

Because , we letu 1 = 1 ln 1 xand Then

0 Exercises 17–20

EXAMPLES 6.6Integrals of type 2 :

(i)

In this case, x is proportional to the derivative of 2 x

2

1 + 1 3: f(x) 1 = 12 x

2

1 + 1 3,

f′(x) 1 = 14 x. Therefore, puttingu 1 = 12 x

2

1 + 13 anddu 1 = 14 x 1 dx,

(ii).

Letu 1 = 1 sin 1 x. Thendu 1 = 1 cos 1 x 1 dx,and

(iii)

Letu 1 = 1 ln 1 x. Then and

0 Exercises 21–26

I

du

u

==+= +Z lnuCln(ln )x C

du

x

= dx

1

,

I

xx

=Z dx

1

ln

.

I

du

u

==+=Z lnuCln(sin )x C+

Ixdx

x

x

==ZZcot dx

cos

sin

I

du

u

==+= ++uC x C

1

4

1

4

1

4

23

2

Z ln ln( )

I

x

x

= dx

• 
  

Z

23

2

.

Z

fx′

fx

dx

()

()

.

IuduuC xC==+= +Z

1

2

1

2

22

(ln )

du

x

= dx

1

.

1

x

d

dx

= lnx

I

x

x

=Z dx

ln