$\underset{0}{\overset{\infty }{\int }}\frac{x.\ln x}{{(1+x^2)}^2}dx=$

$\underset{0}{\overset{\infty }{\int }}\frac{x.\ln x}{{(1+x^2)}^2}dx$

$$\begin{gathered} Put x=\tan \theta \\ dx=se{c}^2\theta d\theta \end{gathered}$$

$$\begin{gathered} =\underset{0}{\overset{\pi /2}{\int }}\frac{\tan \theta .\log \tan \theta }{{(1+{\tan }^2\theta )}^2}se{c}^2\theta d\theta \\ =\underset{0}{\overset{\pi /2}{\int }}\frac{\left(\tan \theta .\log \tan \theta \right)}{se{c}^4\theta }se{c}^2\theta d\theta \\ =\underset{0}{\overset{\pi /2}{\int }}\left(\sin \theta \cos \theta \right)\log \tan \theta d\theta \\ Let I=\frac{1}{2}\underset{0}{\overset{\pi /2}{\int }}\sin 2\theta .\log \tan \theta d\theta \\ I=\frac{1}{2}\underset{0}{\overset{\pi /2}{\int }}\sin 2\left(\frac{\pi }{2}-\theta \right)\log cot\theta d\theta \\ =\frac{1}{2}\underset{0}{\overset{\pi /2}{\int }}\sin 2\theta \log cot\theta d\theta \\ I+I=\frac{1}{2}\underset{0}{\overset{\pi /2}{\int }}\sin 2\theta \left(\log \tan \theta +\log cot\theta \right)d\theta \\ 2I=\frac{1}{2}\underset{0}{\overset{\pi /2}{\int }}\sin 2\theta \left(\log 1\right)d\theta \\ =0 \\ I=0 \end{gathered}$$