The Chemistry Maths Book, Second Edition

6.6 Rational integrands. The method of partial fractions 181

Ifu 1 = 1 f(x) 1 = 1 x

2

1 + 1 px 1 + 1 qthendu 1 = 1 f′(x) 1 dx 1 = 1 (2x 1 + 1 p) 1 dx,and

(6.22)

Therefore,

(6.23)

0 Exercises 66, 67

The numerator is unity:

The integral is evaluated by first transforming the quadratic in the denominator into

the formu

2

1 + 1 a

2

. We write

(6.24)

wherep

2

1 − 14 qis the discriminant of the quadratic. Because it has been assumed that

the quadratic has no real roots, the discriminant is negative, but 4 q 1 − 1 p

2

is positive:

(6.25)

whereu 1 = 1 x 1 + 1 p 22 anda

2

1 = 1 (4q 1 − 1 p

2

) 241 > 10. Then, becausedx 1 = 1 du,

We now use the trigonometric substitution u 1 = 1 a 1 tan 1 θ. Then du 1 = 1 a 1 sec

2

1 θ 1 dθ 1 =

(a 2 cos

2

1 θ) 1 dθ,andu

2

1 + 1 a

2

1 = 1 a

2

(tan

2

1 θ 1 + 1 1) 1 = 1 a

2

2 cos

2

1 θ. Therefore

(6.26)

Whenn 1 = 11 ,

(6.27)

which is one of the standard integrals in Table 6.3. Whenn 1 > 11 , the integral (6.26) can

be evaluated, for example, by reduction as in Example 6.14.

ZZ

du

ua

a

d

a

C

a

u

a

C

22

1

11

• 
  

==+=











+

−

θ

θ

tan

ZZ

du

ua a

d

nn

n

()

cos

22 21

22

1

• 
  

=

−

−

θθ

ZZ

dx

xpxq

du

ua

nn

()()

222

++

=

• 
  

xpxqx

pqp

ua

2

2

2

22

2

4

4

++=+













• 
  

−

















=+

xpxqx

ppq

2

2

2

2

4

4

++=+











 −

−

















Z

dx

xpxq

n

()

2

++

Z

2

1

1

1

2

2

xp

xpxq

dx

xpxqC n

n

n

• 
  

++

=

+++ =

−

−

()

ln( )

()

if

(()xpxq

Cn

21 n

1

++

+>















−

if

ZZ

2

1

1

1

2

xp

xpxq

dx

du

u

uC n

nu

nn

• 
  

++

==

+=

−

−

()

()

ln if

nn

Cn

−

+>















1

if 1