MCQ
$\mathop \smallint \limits_0^1 \frac{{8{\rm{log}}\left( {1 + x} \right)}}{{1 + {x^2}}}dx = $
- A$\frac{\pi }{8}log2$
- B$\;\frac{\pi }{2}log2$
- C$\;log2$
- ✓$\;\pi log2$
✓
Answer
Correct option: D.
$\;\pi log2$
d
$I = \int\limits_0^1 {\frac{{8\log (1 + x)}}{{1 + {x^2}}}dx} $
$I = \int\limits_0^1 {\frac{{8\log (1 + x)}}{{1 + {x^2}}}dx} $
put $x=\tan \theta$
$\therefore \frac{d x}{d \theta}=\sec ^{2} \theta \Rightarrow d x=\sec ^{2} \theta d \theta$
$\therefore I = 8\int\limits_0^{\pi /4} {\frac{{\log (1 + \tan \theta )}}{{1 + {{\tan }^2}\theta }}.} {\sec ^2}\theta d\theta $
$I = 8\int\limits_0^{\pi /4} {\log (1 + \tan \theta )} d\theta .....\left( i \right)$
$ = 8\int\limits_0^{\pi /4} {\log \left[ {1 + \tan \left( {\frac{\pi }{4} - \theta } \right)} \right]} d\theta $
$ = 8\int\limits_0^{\pi /4} {\log \left[ {1 + \frac{{1 - \tan \theta }}{{1 + \tan \theta }}} \right]} d\theta $
$ = 8\int\limits_0^{\pi /4} {\log \left[ {\frac{2}{{1 + \tan \theta }}} \right]} d\theta $
$ = 8\int\limits_0^{\pi /4} {\left[ {\log 2 - \log \left( {1 + \tan \theta } \right)} \right]} d\theta $
$I=8 .(\log 2)[x]_{0}^{\pi / 4}-8$
$ = \int\limits_0^{\pi /4} {\log \left( {1 + \tan \theta } \right)} d\theta $
$I=8 . \frac{\pi}{4} \cdot \log 2-I[\text { From equation }(i)]$
$\Rightarrow 2 I=2 \pi \log 2$
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