_0^1 ^ - 1 x 1 + x^2 ,dx =

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MCQ

$\int_0^1 {\frac{{{{\tan }^{ - 1}}x}}{{1 + {x^2}}}} \,dx = $

A
$\frac{{{\pi ^2}}}{8}$
B
$\frac{{{\pi ^2}}}{{16}}$
C
$\frac{{{\pi ^2}}}{4}$
✓
$\frac{{{\pi ^2}}}{{32}}$

✓

Answer

Correct option: D.

$\frac{{{\pi ^2}}}{{32}}$

d
(d) Put $t = {\tan ^{ - 1}}x$

$\Rightarrow dt = \frac{1}{{1 + {x^2}}}dx,$ then

$\int_0^1 {\frac{{{{\tan }^{ - 1}}x}}{{1 + {x^2}}}dx = \int_0^{\pi /4} {t\,dt = \left[ {\frac{{{t^2}}}{2}} \right]_0^{\pi /4} = \frac{{{\pi ^2}}}{{32}}} } $.

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