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JEE Advanced 2015 Paper 2 Mathematics Q6
Mathematics
Problem
If
α
=
∫
0
1
(
e
9
x
+
3
tan
−
1
x
)
(
12
+
9
x
2
1
+
x
2
)
d
x
,
\alpha=\int_0^1\left(e^{9x+3\tan^{-1}x}\right)\left(\frac{12+9x^2}{1+x^2}\right)\,dx,
α
=
∫
0
1
(
e
9
x
+
3
t
a
n
−
1
x
)
(
1
+
x
2
12
+
9
x
2
)
d
x
,
where
tan
−
1
x
\tan^{-1}x
tan
−
1
x
takes only principal values, then the value of
(
log
e
∣
1
+
α
∣
−
3
π
4
)
\left(\log_e|1+\alpha|-\dfrac{3\pi}{4}\right)
(
lo
g
e
∣1
+
α
∣
−
4
3
π
)
is
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