Solving these equations, we get A = – 5  and B = 10
Thus,
2
2
1
5
6
x
x – x
+
+
 =
5
10
1
2
3
x –
x –
−
+
Therefore,
2
2
1
5
6
x
dx
x – x
+
+
∫
 =
1
5
10
2
3
dx
dx
dx
x –
x –
−
+
∫
∫
∫
= x – 5 log |x – 2| + 10 log |x – 3| + C Example 13 Find 
32
2
(
1) (
3)
x
dx
x
−x
+
+
∫
Solution The integrand is of the type as given in Table 7 2 (4) We write
32
2
(
1) (
3)
x –
x
x
+
+
 =
2
A
B
C
1
3
(
1)
x
x
+x
+
+
+
+
So that
3x – 2 = A (x + 1) (x + 3) + B (x + 3) + C (x + 1)2
= A (x2 + 4x + 3) + B (x + 3) + C (x2 + 2x + 1 )
Comparing coefficient of x 2, x and constant term on both sides, we get
A + C = 0, 4A + B + 2C = 3 and 3A + 3B + C = – 2