180 Chapter 6Methods of integration
EXAMPLE 6.17Integrate.
The integrand can be expressed in terms of partial fractions as
Therefore
EXAMPLE 6.18Integrate.
The cubic in the denominator can be factorized asx
31 − 13 x 1 + 121 = 1 (x 1 − 1 1)
2(x 1 + 1 2)so that
the integrand can be expressed in terms of partial fractions as
Then
0 Exercises 63–65
Integrals of type (ii)
We first consider two special forms.
The numerator is the derivative of the quadratic
In this case, the integral is either of type 2 in Table 6.2 or it is a simple generalization
thereof:
(6.21)
ZZ
2
2xp
xpxq
dx
fx
fx
dx
nn++
=
′
()
()
[()]
=
−
−
−
ln +
x
xx
C
1
2
2
1
=−−
−
ln( )x −++ln( )
x
1 xC
2
1
2
ZZZZ
51
32
1
2
1
2
32x
xx
dx
dx
x
dx
x
dx
x
−+
=
−
−
−
()
51
32
1
1
2
1
1
2
32x
xx
x
x
x
−+
=
−
−
−
()
Z
51
32
3x
xx
dx
−+
=− −+ −
+=
−
−
1
2
24
1
2
4
2
ln()()xxCln ln
x
x
+C
ZZ
dx
xx x x
dx
()()24
1
2
1
2
1
−− 4
=
−
−
−
1
24
1
2
1
2
1
()()−− 4
=
−
−
−
xx x x
Z
dx
()()xx
24 −−