How to solve this integral $\int^{\infty }_{0} {\frac{x \log x}{(1+x^2)^2}} \, dx$? I will be grateful if you would write me a solution procedure for this integral
$$\int^{\infty }_{0} {\frac{x \log x}{(1+x^2)^2}} \, dx. $$
I am sure that an antiderivative is
$$\frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right).$$
Now if I put $+\infty $ instead of $x$ I get
\begin{align*}
\left[ \frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right) \right]^{\infty }_{0}
&= \frac{1}{4} \left( \frac{\infty}{\infty}-\infty \right)-\frac{1}{4} \left( \frac{2 \log 1}{1}-\log 1 \right) \\
&= \frac{1}{4} \left( \frac{\infty}{\infty}-\infty \right).
\end{align*}
As you can see, it is useless. Can you help me please? Thanks
Can I use this solution below? 
Let $$I=\frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right).$$
Now if I calculate the limit of I i get:
$$\lim_{x\to\infty}I=0$$
So the final result is 
\begin{align*}
\left[ \frac{1}{4} \left( \frac{2x^2 \log x}{1+x^2}- \log(1+x^2) \right) \right]^{\infty }_{0}=0
\end{align*}

Putting $x=\frac1y$
$$I=\int_0^{\infty}\frac{x\ln x}{(1+x^2)^2}dx$$
$$=\int_{\infty}^0\frac{(-\ln y)}{y\left(1+\frac1{y^2}\right)^2}\cdot\frac{(-dy)}{y^2}$$
$$=\int_{\infty}^0\frac{\ln ydy}{(1+y^2)^2}$$
$$=-\int_0^{\infty}\frac{\ln dy}{(1+y^2)^2}\text{ as }\int_a^bf(x)dx=-\int_b^af(x)dx$$
$$=-I$$
