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 ? 
 A: Put $x=\dfrac1y$
$$I=\int_\infty^0\dfrac{y^4\cdot -\ln(y)}{y(y^2+1)^2}\cdot-\dfrac{dy}{y^2}=\int_\infty^0\dfrac{y\ln y}{(1+y^2)^2}dy=-\int_0^\infty\dfrac{y\ln y}{(1+y^2)^2}dy=-I$$
A: HINT:
Let $x=\tan y$
$$\implies2J=\int_0^{\pi/2}\sin2y\cdot\ln(\tan y)dy$$
Now use $$I=\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$
$$\implies2I=\int_a^bf(x)dx+\int_a^bf(a+b-x)dx=\int_a^b\left(f(x)+f(a+b-x)\right)dx$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
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 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
\color{#f00}{I} &=
\int_{0}^{\infty}{x\ln\pars{x} \over \pars{1 + x^{2}}^{2}}\,\dd x =
{1 \over 4}\bracks{\overbrace{\int_{0}^{1}{x\ln\pars{x} \over \pars{1 + x}^{2}}\,\dd x}^{\ds{\equiv\ J}}\ +\
\int_{1}^{\infty}{x\ln\pars{x} \over \pars{1 + x}^{2}}\,\dd x}
\end{align}
The second integral in the RHS $\underline{\mbox{is equal to}}$ $\ds{-J}$ after the sub$\ldots$ $\ds{x \to {1 \over x}}$ such that
$\fbox{$\quad\ds{\color{#f00}{I} = J + \pars{-J} = \color{#f00}{0}}\quad$}$.
A: A faster approach:
$$ J(a,b)=\int_{0}^{+\infty}\frac{x^b}{a^2+x^2}\,dx = a^{b-1} \frac{\pi}{2\cos\left(\frac{\pi b}{2}\right)}\qquad (a>0,b\in(-1,1))\tag{1}$$
is a consequence of the reflection formula for the $\Gamma$ function and Euler's beta function properties.
By considering:
$$ \lim_{b\to 1^-}\lim_{a\to 1^-}-\frac{1}{2}\frac{\partial^2 J}{\partial a \partial b} \tag{2}$$
we recover the value of our integral by differentiation under the integral sign. That gives:

$$ I=\int_{0}^{+\infty}\frac{x\log x}{(1+x^2)^2}\,dx = \lim_{b\to 1^-} -\frac{\pi}{8}\cdot \frac{2+\pi(b-1)\tan\left(\frac{\pi b}{2}\right)}{ \cos\left(\frac{\pi b}{2}\right)}=\color{red}{0}\tag{3}$$

that also follows from a symmetry argument:

$$ I = \int_{0}^{1}\frac{x\log x}{(1+x^2)^2}\,dx +\int_{0}^{1}\frac{\frac{1}{x}\log\left(\frac{1}{x}\right)}{x^2\left(1+\frac{1}{x^2}\right)^2}\,dx=\int_{0}^{1}0\,dx = \color{red}{0}.\tag{4} $$

A: 
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!
