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 Then integrating it by
parts, we get
I =
2
1
log (log )
log
(log )
dx
x
x
x dx
x
x
x
−
+
∫
∫
=
2
log (log )
log
(log )
dx
dx
x
x
x
x
−
+
∫
∫ (1)
Again, consider 
log
dx
x
∫
, take 1 as the second function and integrate it by parts,
we have 
2
1
1
–
–
log
log
(log )
dx
x
x
dx
x
x
x
x






= 













∫
∫