Question

Solve ${\int }_0^{\infty }\frac{x\log x}{{(1+x^2)}^2}dx$

Answer 1

$\because I={\int }_0^{\infty }\frac{x\log x}{{(1+x^2)}^2}dx$

Put $x=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta$

$\therefore I={\int }_0^{\frac{\pi }{2}}\frac{\tan \theta \log \left(\tan \theta \right)}{{\sec }^4\theta }.{\sec }^2\theta d\theta$

$={\int }_0^{\frac{\pi }{2}}\frac{\sin \theta \log \left(\tan \theta \right)}{\cos \theta .\frac{1}{{\cos }^2\theta }}d\theta ={\int }_0^{\frac{\pi }{2}}\sin \theta \cos \theta \log \left(\tan \theta \right)d\theta$

$=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\sin 2\theta \log \left(\tan \theta \right)d\theta =0$ $(\because {\int }_0^{\frac{\pi }{2}}\sin 2\theta \log \tan \theta d\theta =0)$