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
=
+ The new limits are, when x = 0, t = 0 and
when x = 1, 
4
t
=π Thus, as x varies from 0 to 1, t varies from 0 to 
4
π Therefore
–1
1
2
0
tan
1
x dx
+x
∫
=
2
4
4
0
t20
t dt
π
π






∫
 = 
2
2
1
– 0
2 16
32


π
=π




EXERCISE 7