A) \[\log 2\]
B) \[\pi \log 2\]
C) \[\frac{\pi }{8}\log 2\]
D) \[\frac{\pi }{2}\log 2\]
Correct Answer: B
Solution :
We have, \[l=\int\limits_{0}^{1}{\frac{8\log \left( 1+x \right)}{1+{{x}^{2}}}dx}\] Put \[x=\tan \theta \] \[\Rightarrow I=\int\limits_{0}^{\frac{\pi }{4}}{8.}\frac{\log \left( 1+\tan \theta \right)}{{{\sec }^{2}}\theta }{{\sec }^{2}}\theta d\theta \] \[=8\int\limits_{0}^{\pi /4}{\log \left( 1+\tan \theta \right)d\theta }\] \[=8\int\limits_{0}^{\pi /4}{\log \left( 1+\frac{10-\tan \theta }{1+\tan \theta } \right)d\theta }\] \[=8\int\limits_{0}^{\pi /4}{\log \left( \frac{2}{1+\tan \theta } \right)d\theta }\] \[=\left( \log 2 \right).\frac{\pi }{4}-I\] \[\Rightarrow 2I=2\pi \log 2\] \[\Rightarrow I=\pi \log 2\]
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