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 Therefore,
2
2
6
5
xx
x
2
++
+
∫
 =
1
4
6
1
4
2
2
6
5
2
6
5
x
dx
dx
x
x
x
x
2
2
+
+
+
+
+
+
∫
∫
=
1
2
1
1
I
I
4
2
+
    (say) (1)
314
MATHEMATICS
In I1, put 2x2 + 6x + 5 = t, so that (4x + 6) dx = dt
Therefore,
I1 =
1
log
C
dt
t
t
=
+
∫
=
2
1
log | 2
6
5|
C
x
x
+
+
+