A) 0
B) 1
C) \[\infty \]
D) None of these
Correct Answer: A
Solution :
\[I=\int_{0}^{\infty }{\frac{x\log x}{{{(1+{{x}^{2}})}^{2}}}\,dx}\] Put \[x=\tan \theta \] Þ \[dx={{\sec }^{2}}\theta \,d\theta \] \[\therefore \]I \[=\int_{0}^{\pi /2}{\frac{\tan \theta \,\log \,(\tan \theta )}{{{\sec }^{4}}\theta }}{{\sec }^{2}}\theta \,d\theta \] \[=\int_{0}^{\pi /2}{\sin \theta \,\cos \theta \,\log \,(\tan \theta )\,d\theta }\] \[=\frac{1}{2}\int_{0}^{\pi /2}{\sin 2\theta \log \,(\,\tan \theta \,)\,d\theta }\]\[=0\], \[\left\{ \because \int_{0}^{\pi /2}{\sin 2\theta \,\,\log \,\,\tan \theta \,\,d\theta =0} \right\}\].
You need to login to perform this action.
You will be redirected in
3 sec
