A) \[-\frac{\pi }{2}\log 2\]
B) \[-\pi \log 2\]
C) \[\pi \log 2\]
D) None of these
Correct Answer: C
Solution :
Let\[I=\int_{0}^{\infty }{\log \left( x+\frac{1}{x} \right)}\frac{dx}{1+{{x}^{2}}}\] \[=\int_{0}^{\infty }{\log \left( \frac{{{x}^{2}}+1}{x} \right)}.\frac{dx}{1+{{x}^{2}}}\] Let \[x=tan\theta \] \[\Rightarrow \] \[dx=se{{c}^{2}}\theta d\theta \] \[\therefore \]\[I=\int_{0}^{\pi /2}{\log \left( \frac{{{\tan }^{2}}\theta +1}{\tan \theta } \right)}\frac{{{\sec }^{2}}\theta d\theta }{1+{{\tan }^{2}}\theta }\] \[=\int_{0}^{\pi /2}{\log \left( \frac{{{\sec }^{2}}\theta }{\tan \theta } \right)}\frac{{{\sec }^{2}}\theta }{{{\sec }^{2}}\theta }d\theta \] \[=\int_{0}^{\pi /2}{(\log }{{\sec }^{2}}\theta -\log \tan \theta )d\theta \] \[=2\int_{0}^{\pi /2}{\log }\sec \theta d\theta -\int_{0}^{\pi /2}{\log }\tan \theta .d\theta \] \[=2\int_{0}^{\pi /2}{\log }\sec \theta d\theta -0\] \[=-2\int_{0}^{\pi /2}{\log }\cos \theta d\theta \] \[\left[ \because \int_{0}^{\pi /2}{\log \tan \theta d\theta =0} \right]\] \[=-2\times -\frac{\pi }{2}\log 2\] \[\left[ \because \int_{0}^{\pi /2}{\log \cos \theta d\theta =-\frac{\pi }{2}\log 2} \right]\] \[=\pi \log 2\]
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