<s>[INST] Use integration by parts $\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$ $$I=\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$$
Clearly $$-2I=\int^{\infty}_{0} \ln x \cdot \frac{-2x }{(1+x^2)^2} dx$$
My attempt :
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \left(\frac{1}{1+x^2}\right) \cdot \frac{1}{x} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1}{x(1+x^2)}  dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1+x^2-x^2}{x(1+x^2)}  dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1}{x}+ \frac{1}{2}\int^{\infty}_{0} \frac{2x}{1+x^2} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 -\left[ \ln x -\frac{1}{2}\cdot \ln(1+x^2) \right]^\infty_0 $$
$$-2I=\left[  \frac{\ln x}{1+x^2}\right]^\infty_0 -\left[\ln \left (\frac{x}{\sqrt{1+x^2}} \right) \right]^\infty_0 $$
How can I evaluate the last limits ? 
 [/INST] 
Integration by parts works fine provided one writes the integral $I$ as
$$I=\lim_{\epsilon\to 0^+}\lim_{L\to \infty}\int_{\epsilon}^L\frac{x\log(x)}{(1+x^2)^2}\,dx$$

Then, proceeding with integration by parts, we find 
$$\begin{align}
I&=\frac14\lim_{\epsilon\to 0^+}\lim_{L\to \infty}\left.\left(\frac{2x^2\log(x)}{1+x^2}-\log(1+x^2)\right)\right|_{\epsilon}^L\\\\
&=\frac14\lim_{L\to \infty}\left(\frac{2L^2\log(L)}{1+L^2}-\log(1+L^2)\right)\\\\
&-\frac14\lim_{\epsilon\to 0^+}\left(\frac{2\epsilon^2\log(\epsilon)}{1+\epsilon^2}-\log(1+\epsilon^2)\right)\tag 1
\end{align}$$
Now, since $\lim_{\epsilon\to 0^+}\epsilon^2 \log(\epsilon)=0$, the contribution from the evaluation of the lower limit vanishes.
To evaluate the contribution from the upper limit, we write
$$\begin{align}
\frac{2L^2\log(L)}{1+L^2}-\log(1+L^2)&=\frac{2L^2\log(L)}{1+L^2}-2\log(L)-\log(1+1/L^2)\\\\
&=-2\frac{\log(L)}{1+L^2}-\log(1+1/L^2)
\end{align}$$
which clearly approaches zero as $L\to \infty$.
And that is that!
</s>