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*}
 A: Putting $x=\tan\theta$  we get,
$$I=\int_0^{\frac\pi2}\frac{\tan\theta\ln \tan \theta}{\sec^2\theta}d\theta$$
$$=\frac12\int_0^{\frac\pi2}\sin2\theta\ln \tan \theta d\theta$$
Now as $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$
$$I=\frac12\int_0^{\frac\pi2}\sin2\left(\frac\pi2+0-\theta\right)\theta\ln \tan \left(\frac\pi2+0-\theta\right) d\theta$$
$$=-\int_0^{\frac\pi2}\sin2\theta \ln\tan \theta d\theta=-I$$
as $\sin(\pi-2\theta)=\sin2\theta$ 
and $\tan \left(\frac\pi2+0-\theta\right)=\cot\theta=(\tan\theta)^{-1}\implies \ln \tan\left(\frac\pi2+0-\theta\right)=-\ln\tan \theta$
$$\implies I=0$$
A: Putting $x=\frac1y$
$$I=\int_0^{\infty}\frac{x\ln x}{(1+x^2)^2}dx$$
$$=\int_{\infty}^0\frac{(-\ln y)}{y\left(1+\frac1{y^2}\right)^2}\cdot\frac{(-dy)}{y^2}$$
$$=\int_{\infty}^0\frac{\ln ydy}{(1+y^2)^2}$$
$$=-\int_0^{\infty}\frac{\ln dy}{(1+y^2)^2}\text{ as }\int_a^bf(x)dx=-\int_b^af(x)dx$$
$$=-I$$
A: Related problems: (I). Recalling the Mellin transform
$$ F(s) = \int_{0}^{\infty} x^{s-1}f(x) dx \implies  F'(s) = \int_{0}^{\infty} x^{s-1}\ln(x)f(x) dx$$
So, taking $f(x)=\frac{1}{(1+x^2)^2}$ and finding its Mellin transform
$$F(s) = \frac{1}{4}\,{\frac { \left( 2-s \right) \pi }{\sin \left( \frac{\pi s}{2}
 \right) }}.$$
Now, differentiating and taking the limit as $s\to 2$ gives the desired result
$$ \lim_{s\to 2} F'(s) = 0. $$
A: As $x \to \infty$, 
$$
\frac{2x^2\log(x)}{1+x^2} - \log(1+x^2) \approx \frac{2x^2\log(x)}{x^2} - \log(x^2) = 0.
$$
You can make the $\approx$ precise using explicit estimates.
