A) \[\frac{\pi }{2}\]
B) 0
C) 1
D) \[\frac{\pi }{4}\]
Correct Answer: D
Solution :
\[\int_{0}^{\infty }{\frac{x\,d\,x}{(1+x)\,(1+{{x}^{2}})}}\] Put \[x=\tan \theta \] \[\Rightarrow \] \[\frac{dx}{d\theta }={{\sec }^{2}}\theta \] \[=\int_{0}^{\pi /2}{\frac{\tan \theta {{\sec }^{2}}\theta d\theta }{(1+\tan \theta )\,(1+{{\tan }^{2}}\theta )}}\] [For limit, \[x=0\Rightarrow \theta =0,\,x=\infty \Rightarrow \theta =\pi /2\]] \[=\int_{0}^{\pi /2}{\frac{\tan \theta {{\sec }^{2}}\theta d\theta }{(1+\tan \theta ){{\sec }^{2}}\theta }}\] \[[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta ]\] \[=\int_{0}^{\pi /2}{\frac{\tan \theta }{1+\tan \theta }}\] \[=\int_{0}^{\pi /2}{\frac{\frac{\sin \theta }{\cos \theta }}{1+\frac{\sin \theta }{\cos \theta }}d\theta }\] \[=\int_{0}^{\pi /2}{\frac{\sin \theta }{\cos \theta +\sin \theta }d\theta }\] Let \[I=\int_{0}^{\pi /2}{\frac{\sin \theta }{\cos \theta +\sin \theta }d\theta }\] ... (1) \[I=\int_{0}^{\pi /2}{\frac{\sin \left( \frac{\pi }{2}-\theta \right)}{\cos (\pi /2-\theta +\sin \left( \frac{\pi }{2}-\theta \right)}}\] \[\left[ \int_{0}^{a}{f(x)\,dx=\int_{0}^{a}{f(a-ax)\,dx}} \right]\] \[I=\int_{0}^{\pi /2}{\frac{\cos \theta }{\sin \theta +\cos \theta }}\] ?. (2) \[[\sin \,(\pi /2-\theta )=\cos \theta ,\,\cos \left( \frac{\pi }{2}-\theta \right)=\sin \theta ]\] adding (1) and (2) we get \[2I=\int_{0}^{\pi /2}{\frac{\sin \theta }{\sin \theta +cos\theta }d\theta +}\int_{0}^{\pi /2}{\frac{\cos \theta }{\sin \theta +\cos \theta }d\theta }\] \[2I=\int_{0}^{\pi /2}{\frac{\sin \theta +\cos \theta }{\sin \theta +\cos \theta }d\theta }\] \[2I=\int_{0}^{\pi /2}{d\theta }\] \[2I=\left[ \begin{align} & 0 \\ & \\ \end{align} \right]_{0}^{\frac{\pi }{2}}\] \[2I=\frac{\pi }{2}\] \[\Rightarrow \] \[I=\frac{\pi }{4}\]
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