To test convergence of improper integral $ \int_{0}^{\infty} \frac{x\log(x)}{(1+x^2)^2}\, \mathrm dx$ I have to test convergence of improper integral
$$ \int_{0}^{\infty} \frac{x\log(x)}{(1+x^2)^2}\,\mathrm dx$$
I write as $\log(x) \leq x$ . So $x\log(x) \leq x^2$. So $ \frac{x\log(x)}{(1+x^2)^2} \leq \frac{x^2}{(1+x^2)^2}$ . Now using comparison test with integral  $\frac{1}{x^2}$ to get original integral convergent. Is that right ? Not sure Thanks
 A: This is probably not an answer but it is too long for a comment.
$$I=\int_{0}^{\infty} \frac{x\log(x)}{(1+x^2)^2}\, dx=\int_{0}^{1} \frac{x\log(x)}{(1+x^2)^2}\, dx+\int_{1}^{\infty} \frac{x\log(x)}{(1+x^2)^2}\, dx$$ As  lab bhattacharjee commented, change variable $x=\frac 1y$ for the second integral; so  $$\int_{1}^{\infty} \frac{x\log(x)}{(1+x^2)^2}\, dx=\int_1^0 \frac{y\log(y)}{(1+y^2)^2}\, dy=-\int_0^1 \frac{y\log(y)}{(1+y^2)^2}\, dy$$ which makes the beautiful result $I=0$.
A: I will show that the integral converges by actually evaluating it. Firstly, consider the following complex contour integral:
$$J:=\oint_C \frac{z \ln^2 z}{\left ( 1+z^2 \right )^2}dz$$
where $C$ is a dumbell contour pictured here, with the branch cut along the positive part of the real axis:

We see that $$J=\int _{c1}+\int _{c2}+\int_{C_{R}} +\int _{C \epsilon}$$
I want to show that the integral along the outer circle of radius R tends to zero for large R. Recall the estimation lemma:
$$\left | \int _{C_R}f\left ( z \right ) dz\right |\leq 2 \pi \frac{R^2\left ( \ln^2 R+\theta \right )}{\left ( R-1 \right )^4}, \theta \in \left [0, 2 \pi  \right )$$ 
Taking the limit as $R\rightarrow  \infty$ and knowing that $ \ln x  \leq x$ for large $x$ we get that $\left | \int _{C_R}f\left ( z \right ) dz\right |\rightarrow 0$
Similarly, we show that $\left | \int _{C\epsilon}f\left ( z \right ) dz\right |\rightarrow 0$, acknowledging the fact that $x \ln x \rightarrow 0$ as $x \rightarrow 0$.
Now we are left with only two integrals, along C1 and along C2.
$$\int _{C1}f\left ( z \right )dz=\int_{0}^{\infty}\frac{x \ln ^2 x}{\left ( 1+x^2 \right )^2}dx$$
The argument of the log has a phase of $2 \pi i$ along C2, so :
$$\int _{C2}f\left ( z \right )dz=\int_{0}^{\infty}\frac{x \left ( \ln x +i2 \pi \right )^2}{\left ( 1+x^2 \right )^2}dx$$
Adding them together leaves us with :
$$J:=-4 i \pi\int_{0}^{\infty}\frac{x \ln x}{\left ( 1+x^2 \right )^2}dx+4 \pi^2\int_{0}^{\infty}\frac{x dx}{\left ( 1+x^2 \right )^2} (*)$$
On the other hand $$J:=2 \pi i \sum Res f\left ( z \right )$$
The function has poles of order 2, at $z= \pm i$, so
$$J:=2 \pi i \left ( i\frac{\pi}{4} -i\frac{\pi}{4}\right )=0(**)$$
By equating $(*)$ and $(**)$ we get that
$$\int_{0}^{\infty}\frac{x \ln x}{\left ( 1+x^2 \right )^2}dx=0$$
and also $$\int_{0}^{\infty}\frac{x}{\left ( 1+x^2 \right )^2}dx=0$$
A: Letting $u=x^2+1$ then $\frac{1}{2}\,\mathrm du=x\,\mathrm dx$ and $\log(x)=\log((u-1)^\frac{1}{2})=\frac{1}{2}\log(u-1)$.  With that substitution I think the convergence is easier to see.  Leibovici's answer is the most beautiful answer, imo.
