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