here is the integral:
$$\int_0^{+\infty} \frac{x\mathrm{ln}x}{(1+x^2)^2}dx$$
What is the best way to solve it? I was thinking of by parts but I did not find exact result then, if there is substitution in the denum, it does not "cover" $\mathrm{ln}x$ and I don't get more-less table integral (or similar).
Are there easier ways?
On
$$\int_{1}^{+\infty}\frac{x\log x}{(1+x^2)^2}\,dx \stackrel{x\mapsto \frac{1}{z}}{=} -\int_{0}^{1}\frac{z \log z}{(1+z^2)^2}\,dz$$
hence the value of the integral is simply $\color{red}{\large 0}$.
The absolute convergence of the original integral follows from
$$ \forall x>0,\qquad \left|x\log x\right|\leq \max(1,x\sqrt{x}).$$
As an addendum, it might be interesting to know what is the area that gets "canceled out", i.e. what is the value of the integral $$ I_1=\int_{0}^{1}\frac{z}{(1+z^2)^2}(-\log z)\,dz \stackrel{\text{IBP}}{=}\frac{1}{2}\int_{0}^{1}\frac{z}{1+z^2}\,dz=\color{red}{\frac{\log 2}{4}}. $$ Integration by parts (IBP) here is performed by taking $\frac{1}{2}\left(1-\frac{1}{1+z^2}\right)=\frac{z}{2(1+z^2)}$ as a primitive of $\frac{z}{(1+z^2)^2}$, in order to ensure the existence of $\lim_{x\to 0^+}\frac{1}{2}\left(1-\frac{1}{1+z^2}\right)(-\log z)$.
On
hint
By parts, it becomes
$$\Bigl[-\frac {\ln (x)}{2(1+x^2)}\Bigr]_X^{+\infty}+\int_X^{+\infty}\frac {dx}{2x (1+x^2)}$$
with $t=x^2$, we find at the end
$$\frac {\ln (X)}{2 (1+X^2)}-\frac {1}{2}\ln (\frac {X}{1+X^2})$$
$$=-\frac {X^2\ln (X)}{1+X^2}+\frac {\ln (1+X^2)}{2} $$
Now compute the limit when $X\to 0^+$ to find zero.
HINT:
Enforce the substitution $x\to 1/x$ and watch the "magic" ensue.
If you would like to proceed less efficiently, integrate by parts with $u=\log(x)$ and $v=\frac{-1/2}{1+x^2}$. Then, use partial fraction expansion to write $\frac{1}{x(x^2+1)}=\frac1x-\frac x{x^2+1}$. Be mindful that the integral is improper and that evaluation of the terms involved in the integration by parts scheme do *not * individually converge. Rather, their sum converges.