Therefore,
4
5
5
1
x
x
dx
+
∫
 =
∫t dt
 = 
223
3
t  = 
3
5
2
2 (
1)
3
x +
Hence,
1
4
5
15
1
x
x
dx
−
+
∫
 =
1
3
5
2
– 1
2 (
1)
3
x


+






=
(
)
3
3
5
5
2
2
2 (1
1) – (–1)
1
3


+
+






=
3
3
2
2
2 2
0
3


−






 = 
2
4
2
(2
2)
3
3
=
Alternatively, first we transform the integral and then evaluate the transformed integral
with new limits 340
MATHEMATICS
Let
t = x5 + 1 Then dt = 5 x4 dx Note that, when
x = – 1, t = 0 and when x = 1, t = 2
Thus,  as x varies from – 1 to 1, t varies from 0 to 2
Therefore
1
4
5
15
1
x
x
dx
−
+
∫
 =
2
0
t dt
∫
=
2
3
3
3
2
2
2
0
2
2 2 – 0
3
3
t



=












 = 
2
4
2
(2
2)
3
3
=
Example 29 Evaluate 
– 1
1
2
0
tan
1
x dx
+x
∫
Solution Let t = tan – 1x, then 
12
1
dt
dx
x
=
+