A) \[\pi \log 2\]
B) \[-\pi \log 2\]
C) \[(\pi /2)\log 2\]
D) \[-(\pi /2)\log 2\]
Correct Answer: A
Solution :
\[I=\int_{0}^{\infty }{\log \left( x+\frac{1}{x} \right)}\frac{1}{1+{{x}^{2}}}dx\] Put \[x=\tan \theta \Rightarrow \,\,dx={{\sec }^{2}}\theta \,\,d\theta \] \[\Rightarrow I=\int_{0}^{\pi /2}{\,\,\,\,\,\log (\tan \theta +\cot \theta })\frac{{{\sec }^{2}}\theta }{{{\sec }^{2}}\theta }\,d\theta \] Þ \[I=\int_{0}^{\pi /2}{\,\,\,\,\,\log (\tan \theta +\cot \theta })d\theta \] \[\Rightarrow I=\int_{0}^{\pi /2}{\log \frac{(1+{{\tan }^{2}}\theta )}{\tan \theta }\,d\theta }\] Þ I \[=2\int_{0}^{\pi /2}{\log \sec \theta \,d\theta -\int_{0}^{\pi /2}{\log \tan \theta }}\,d\theta \] Þ I \[=2\int_{0}^{\pi /2}{\log \sec \theta \,\,d\,\theta }\]; \[\left\{ \,\because \int_{0}^{\pi /2}{\log \tan \theta =0} \right\}\] \[\Rightarrow \,I=-2\int_{0}^{\pi /2}{\,\,\,\,\,\log \cos \theta \,d\theta }\] Þ\[I=-2\times \frac{-\pi }{2}\log 2\], \[\left\{ \because \int_{0}^{\pi /2}{\log \cos \theta =-\frac{\pi }{2}\log 2} \right\}\] Þ \[I=\pi \log 2\].
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