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 Solving these equations, we get
11
5
11
A
B
and C
4
2
4
–
–
,
=
=
= Thus the integrand is given by
32
2
(
1) (
3)
x
x
x
−
+
+
 =
2
11
5
11
4(
1)
4(
3)
2(
1)
–
–
x
x
x
+
+
+
Therefore,
32
2
(
1) (
3)
x
x
x
−
+
+
∫
 =
2
11
5
11
4
1
2
4
3
(
1)
dx
dx
dx
–
x
x
x
−
+
+
+
∫
∫
∫
=
11
5
11
log
+1
log
3
C
4
2( + 1)
4
x
x
x
+
−
+
+
= 11
+1
5
log
+C
4
+3
2 ( +1)
xx
x
+
320
MATHEMATICS
Example 14 Find 
2
2
2
(
1) (
4)
x
dx
x
x
+
+
∫
Solution  Consider 
2
2
2
(
1) (
4)
x
x
x
+
+
 and put x2 = y Then
2
2
2
(
1) (
4)
x
x
x
+
+
 = (
1) (
4)
y
y
y
+
+
Write
(
1) (
4)
y
y
y
+
+
 =
A
B
1
4
y
y
+
+
+
So that
y =  A (y + 4) + B (y + 1)
Comparing coefficients of y and constant terms on both sides, we get A + B = 1
and 4A + B = 0, which give
A =
1
4
and B
3
3
−
=
Thus,
2
2
2
(
1) (
4)
x
x
x
+
+
 =
2
2
1
4
3(
1)
3 (
4)
–
x
x
+
+
+
Therefore,
2
2
2
(
1) (
4)
x dx
x
x
+
+
∫
 =
2
2
1
4
3
3
1
4
dx
dx
–
x
x
+
+
+
∫
∫
=
1
1
1
4
1
tan
tan
C
3
3
2
2
–
– x
–
x +
×
+
=
1
1
1
2
tan
tan
C
3
3
2
–
– x
–
x +
+
In the above example, the substitution was made only for the partial fraction part
and not for the integration part