Prove that $\int_0^\infty \frac{x\,\log x}{(1+x^2)^2} = 0$ I have been asked to prove that $$\int_0^\infty \frac{x\,\log x}{(1+x^2)^2}\,dx = 0$$
I tried proceeding as follows:
\begin{align}
\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 \\
& = \int_0^1 \frac{x\,\log x}{(1+x^2)^2}\,dx + \int_{u=1}^0 \frac{u\,\log u}{(1+u^2)^2}\,du\quad (\text{where }u=1/x) \\
&= \int_0^1 \frac{x\,\log x}{(1+x^2)^2}\,dx - \int_0^1 \frac{u\,\log u}{(1+u^2)^2}\,du \\
&=^? \int_0^1 0 \,dx \\
&= 0
\end{align}
But I think that I should first show that the improper integral $\int_0^1 {x\,\log x}/{(1+x^2)^2}\,dx$ is convergent.
Do I need to show it? If yes, how to do so?
(I have been taught basic methods such as comparison, limit comparison and Dirichlet test.)

$I$ converges because 
$$\dfrac{x\ln x}{(1+x^2)^2}\le \dfrac{x^2}{(1+x^2)^2}\le\dfrac{1}{1+x^2} ~~~x>1$$ 
and $$ \lim_{x\to 0}\dfrac{x\ln x}{(1+x^2)^2} = 0$$
that is $$x\mapsto \dfrac{x\ln x}{(1+x^2)^2}~~~\text{is continuous on [0,1]}$$
Now Set $x=\frac{1}{u}$ that is $dx = -\frac{1}{u^2}$
$$I=\int_0^\infty\dfrac{x\ln x}{(1+x^2)^2}dx=-\int^0_\infty\dfrac{\frac{1}{u}\ln \frac{1}{u}}{(1+\frac{1}{u^2})^2}\frac{1}{u^2}du =\int_0^\infty\dfrac{\ln \frac{1}{u}}{(\frac{u^2+1}{u^2})^2}\frac{1}{u^3}du$$
But $$\color{red}{ \ln\frac{1}{u} = -\ln u}$$
hence We have 
$$I=\int_0^\infty\dfrac{x\ln x}{(1+x^2)^2}dx\\=\int_0^\infty\dfrac{\ln \frac{1}{u}}{(\frac{u^2+1}{u^2})^2}\frac{1}{u^3}du\\= -\int_0^\infty\dfrac{u^4\ln u}{(u^2+1)^2}\frac{1}{u^3}du \\=-\int_0^\infty\dfrac{u\ln u}{(u^2+1)^2}du  =-I$$
That $$\color{blue}{ I = -I\implies 2I= 0 \implies I= 0}$$
