Therefore,
3
13
10
dx
x
x
2 +
−
∫
 =
2
2
31
17
6
dt
t


− 



∫
=
1
17
1
6
log
C
17
17
3
2
6
6
t –
t
+
× ×
+
[by 7 4 (i)]
=
1
13
17
1
6
6
log
C
13
17
17
6
6
x
–
x
+
+
+
+
=
1
1
6
4
log
C
17
6
xx30
−
+
+
=
1
1
3
2
1
1
log
C
log
17
5
17
3
xx
−
+
+
+
=
1
3
2
log
C
17
xx5
−
+
+
, where C = 
1
1
1
C
17log
3
+
INTEGRALS         313
(iii)
We have 
2
2
5
2
5
5
dx
dx
x
x
x
x –
2
=


−




∫
∫
=
2
2
1
5
1
1
5
5
dx
x –
–












∫
 (completing the square)
Put 
1
5
x –
t
= Then dx = dt Therefore,
5
2
dx
x
x
2 −
∫
 =
2
2
1
5
1
5
dt
t –





∫
=
2
2
1
1
log
C
5
5
t
t – 

+
+




[by 7