The Chemistry Maths Book, Second Edition

6.4 Integration by parts 175

In general, a polynomial of degree ncan be removed by nsuccessive integrations by

parts. The exception to the rule is when the other factor is a logarithmic function.

EXAMPLE 6.12Integrate by parts.

In this case, choosing u 1 = 1 x

n

leads to a more complicated integral. The correct

choice isu 1 = 1 ln 1 xand Then andv 1 = 1 x

n+ 1

2 (n 1 + 1 1),and

A special case of this integral is

0 Exercises 46–48

Integration by parts is straightforward only if one of the factors is a polynomial.

EXAMPLE 6.13Integrate.

In this case either factor can be chosen as uin (6.14); for example, ifu 1 = 1 e

−ax

and

then, for the indefinite integral,

= 1 e

−ax

1 sin 1 x 1 − 1 ae

−ax

1 cos 1 x 1 − 1 a

2

I

=− −

−− −

e x ae x a e x dx

ax ax ax

sin cos cos

2

Z

=+−−

















−−−

exae xaexdx

ax ax ax

sin cos Z cos

Ie xdxe xae xdx

ax ax ax

==+

−−−

ZZcos sin sin

d

dx

x

v

=cos

Z

0

0

∞

exdxa

−ax

cos , ( > )

Zlnxdxx xxC=−+ln

=

• 
  

+−













• 
  

+

1

1

11

2

1

()

()ln

n

xn x C

n

=

• 
  

−

• 
  

• 
  

++

1

1

1

1

1

2

1

n

xx

n

xC

nn

ln

()

=

• 
  

−

• 
  

+

1

1

1

1

1

n

xx

n

xdx

nn

ln Z

ZZxxdx

n

xx

n

x

x

dx

nn n

ln = ln

• 
  

−

• 
  

×

++

1

1

1

1

1

11

du

dx x

=

d 1

dx

x

n

v

=.

Zxxdxn

n

ln , ( ≠− 1 )