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) Let us express
x + 3 = 
2
A
(5
4
) + B
d
– x – x
dx
= A (– 4 – 2x) + B
Equating the coefficients of x and the constant terms from both sides, we get
– 2A = 1 and – 4 A + B = 3, i