<s>[INST] 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*}
 [/INST] 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.
</s>