JEE-Advanced 2015 (26/40)
If :
α
=
∫
0
1
(
e
9
x
+
3
tan
−
1
x
)
(
1
+
x
2
1
2
+
9
x
2
)
d
x
where
tan
−
1
x
takes only principal values, then find the value of
(
lo
g
e
∣
1
+
α
∣
−
4
3
π
)
The answer is 9.
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1 solution
No need of substitution,
Given integral is of the form,
∫
e
g
(
x
)
(
g
′
(
x
)
f
(
x
)
+
f
′
(
x
)
)
=
e
g
(
x
)
f
(
x
)
Here ,
g
(
x
)
=
9
x
+
3
tan
−
1
x
;
f
(
x
)
=
1