Let us illustrate this by examples Example 28 Evaluate 
1
4
5
15
1
x
x
dx
−
+
∫ Solution Put  t = x5 + 1, then dt = 5x4 dx 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