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 4 (4)]
=
2
1
1
2
log
C
5
5
5
x
x –
x –
+
+
Example 10 Find the following integrals:
(i)
2
2
6
5
x
dx
x
x
2
++
+
∫
(ii)
2
3
5
4
x
dx
x
x
+
−
+
∫
Solution
(i)
Using the formula 7 4 (9), we express
x + 2 = 
(
)
2
A
2
6
5
B
d
x
x
dx
+
+
+
 = A (4
6)
B
x +
+
Equating the coefficients of x and the constant terms from both sides, we get
4A = 1 and 6A + B = 2   or    A = 1
4  and B = 1
2