(1)
Now express
2
1
(
1)(
1)
x
x
−
+
 =
2
A
B
C
(
1)
(
1)
x
x
x
+
+
−
+ (2)
INTEGRALS         349
So
1 = A (x2 + 1) + (Bx + C) (x – 1)
= (A + B) x2 + (C – B) x + A – C
Equating coefficients on both sides, we get A + B = 0, C – B = 0 and A – C = 1,
which give 
1
1
A
, B
C
–
2
2
=
=
= Substituting values of A, B and C in (2), we get
2
1
(
1) (
1)
x
x
−
+
 =
2
2
1
1
1
2(
1)
2 (
1)
2(
1)
x
x
x
x
−
−
−
+
+ (3)
Again, substituting (3) in (1), we have
4
2
(
1) (
1)
x
x
x
x
−
+
+
 =
2
2
1
1
1
(
1)
2(
1)
2 (
1)
2(
1)
x
x
x
x
x
+
+
−
−
−
+
+
Therefore
4
2
2
– 1
2
1
1
1
log
1 –
log (
1) –
tan
C
2
2
4
2
(
1) (
1)
x
x
dx
x
x
x
x
x
x
x
=
+
+
−
+
+
−
+
+
∫
Example 40 Find 
2
1
log (log )
(log )
x
dx
x


+




∫
Solution Let 
2
1
I
log (log )
(log )
x
dx
x


=
+




∫
=
2
1
log (log )
(log )
x dx
dx
x
+
∫
∫
In the first integral, let us take 1 as the second function