(2)
and
I2 =
2
2
1
5
2
2
6
5
3
2
dx
dx
x
x
x
x
=
+
+
+
+
∫
∫
=
2
2
21
3
1
2
2
dx
x



+
+








∫
Put 
3
2
x
t
+
= , so that dx = dt, we get
I2 =
2
2
21
1
2
dt
t


+ 



∫
 = 
1
2
1
tan 2
C
1
2
2
–
t +
×
[by 7 4 (3)]
=
1
2
3
tan 2
+ C
2
–
x

+



 = 
(
)
1
2
tan
2
3 + C
–
x + (3)
Using (2) and (3) in (1), we get
(
)
2
1
2
1
1
log 2
6
5
tan
2
3
C
4
2
2
6
5
–
x
dx
x
x
x
x
x
2
+
=
+
+
+
+
+
+
+
∫
where,
C =
1
2
C
C
4
2
+
(ii)
This integral is of the form given in 7 4 (10)