The value of the integral $\underset{0}{\overset{\infty }{\int }}\frac{x\log x}{{(1+x^2)}^2}dx=$

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

Put $x=\tan \theta$

$\begin{array}{l}dx={\sec }^2\theta \cdot d\theta \\ x=0,\infty \Rightarrow \theta =0,\frac{\pi }{2}\\ \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\tan \theta \cdot \mathrm{logtan}\theta }{{(1+{\tan }^2\theta )}^2}\cdot {\sec }^2\theta \cdot d\theta \\ =\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\tan \theta \cdot \mathrm{logtan}\theta }{{\sec }^2\theta }\cdot d\theta \\ I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\sin \theta \cdot \cos \theta \cdot \mathrm{logtan}\theta \to \left(1\right)\\ I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \theta \cdot \sin \theta \cdot \mathrm{logcot}\theta \to \left(2\right)\end{array}$

King’s rule :

$\begin{array}{l}\underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx\\ \left(1\right)+\left(2\right)\\ \Rightarrow 2I=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \theta \cdot \sin \theta (\mathrm{logtan}\theta +\mathrm{logcot}\theta )\cdot d\theta \\ =\underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \theta \cdot \sin \theta \left(0\right)\cdot d\theta \\ =0\\ \Rightarrow I=0\end{array}$