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If
α
=
∫
0
1
(
e
9
x
+
3
tan
−
1
x
)
(
12
+
9
x
2
1
+
x
2
)
d
x
\alpha = \int_{0}^{1} (e^{9x+3\tan^{-1}x}) \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
,…
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} (e^{9x+3\tan^{-1}x}) \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 value of
(
log
e
∣
1
+
α
∣
−
3
π
4
)
\left( \log_e |1+\alpha| - \frac{3\pi}{4} \right)
(
lo
g
e
∣1
+
α
∣
−
4
3
π
)
is [Integer Answer Type 2015]
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