MCQ
$\int_0^1 {\frac{{{{\tan }^{ - 1}}x}}{{1 + {x^2}}}} \,dx = $
- A$\frac{{{\pi ^2}}}{8}$
- B$\frac{{{\pi ^2}}}{{16}}$
- C$\frac{{{\pi ^2}}}{4}$
- ✓$\frac{{{\pi ^2}}}{{32}}$
$\Rightarrow dt = \frac{1}{{1 + {x^2}}}dx,$ 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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