Ways to prove that $\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x = 0$. I have managed to solve it in one way, but I became very interested in this failed attempt.
$$
\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x 
 = \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x
- \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x
$$
We only have to show that those two on the right are equal. And numerical evaluations seem to suggest that they both are in fact $-\frac{\pi}{4}$ but I don't know how to break these down.

I am currently really interested in proving this
  $$ \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x = \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x = -\frac{\pi}{4} $$

Anyway, here's my trivial solution using $u = \frac1x$:
$$
\begin{align}
\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x & = \int_0^1 \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x + \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = \int_\infty^1 \frac{\frac{1}{u^2} \ln(\frac1u)}{(1+\frac{1}{u^2})^3} \frac{-1}{u^2} {\rm d}u
+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = -\int_1^\infty \frac{\ln(u)}{u(u+\frac{1}{u})^3} {\rm d}u 
+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = - \int_1^\infty \frac{u^2 \ln(u)}{(1+u^2)^3} {\rm d}u
+ \int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x \\
& = 0
\end{align}
$$
I'm sure there are many more interesting methods for cracking this integral, since it's so closely related to the popular $\int_0^\infty \frac{\ln(x)}{1+x^2} {\rm d}x = 0$. Please share them if you do come up with any.
 A: $$\int_{0}^{+\infty}\frac{x^{2+\alpha}}{(1+x^2)^3}\,dx \stackrel{(*)}{=}\frac{\pi(1-\alpha^2)}{16\cos\frac{\pi \alpha}{2}} $$
$(*)$: we use the substitution $\frac{1}{1+x^2}=u$, the Beta function and the reflection formula for the $\Gamma$ function. 
This holds for any $\alpha$ such that $-3<\text{Re}(\alpha)<3$, and since the RHS is an even function, the origin is a stationary point, i.e.
$$\color{red}{0}=\frac{d}{d\alpha}\left.\int_{0}^{+\infty}\frac{x^{2+\alpha}}{(1+x^2)^3}\,dx\right|_{\alpha=0}\stackrel{\text{DCT}}{=}\int_{0}^{+\infty}\frac{x^{2}\log x}{(1+x^2)^3}\,dx.$$
A: I found a way to directly evaluate both of those integrals!
We first use the well known result from $\int_0^{\infty}\frac{\ln x}{x^2+a^2}\mathrm{d}x$ Evaluate Integral
$$
\int_0^\infty \frac{\ln(x)}{x^2 +\alpha^2} {\rm d}x = \frac{\pi}{2\alpha} \ln(\alpha)
$$
and use differentiation under the integral sign.
$$
\int_0^\infty \frac{-2\alpha\ln(x)}{(x^2 +\alpha^2)^2} {\rm d}x
= \frac{\pi}{2} \frac{1-\ln(\alpha)}{\alpha^2} \\
\implies \int_0^\infty \frac{\ln(x)}{(x^2 +\alpha^2)^2} {\rm d}x
= \frac{\pi}{4\alpha^3} (\ln(\alpha)-1)
$$
And again, to get
$$
\int_0^\infty \frac{\ln(x)}{(x^2 +\alpha^2)^3} {\rm d}x
= \frac{\pi}{16\alpha^5}(3\ln(\alpha) - 4)
$$

And so setting $\alpha = 1$ gives us the immediate result
  $$  \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x = \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x = -\frac{\pi}{4} $$

Another method to evaluate this is to use the integral from the 2015 MIT Integration Bee, and is also how I first came across this integral.
From this result (solved by using the substitution $u=\frac1x$)
$$ \int_0^\infty \frac{1}{(1+x^2)(1+x^\alpha)} {\rm d}x = \frac\pi4 $$
we will get, by differentiating with respect to $\alpha$,

$$ \int_0^\infty \frac{x^\alpha \ln(x)}{(1+x^2)(1+x^\alpha)^2} {\rm d}x = 0 $$

And finally setting $\alpha = 2$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{\Re\pars{\mu} > - 1}$ and
  $\ds{\Re\pars{\nu} > 0}$:

\begin{align}
I_{\mu\nu} & \equiv
\bbox[10px,#ffd]{\int_{0}^{\infty}{x^{\mu}\ln\pars{x} \over \pars{1 + x^{2}}^{\nu}}\,\dd x}
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
{1 \over 4}\int_{0}^{\infty}
{x^{\mu/2 - 1/2}\ln\pars{x} \over \pars{1 + x}^{\nu}}\,\dd x
\\[5mm] & =
\left.{1 \over 4}\,\partiald{}{\alpha}\int_{0}^{\infty}
{x^{\alpha + \mu/2 - 1/2} \over \pars{1 + x}^{\nu}}\,\dd x
\,\right\vert_{\ \alpha\ =\ 0}
\\[5mm] & \stackrel{x + 1\ \mapsto\ x}{=}\,\,\,
\left.{1 \over 4}\,\partiald{}{\alpha}\int_{1}^{\infty}
{\pars{x - 1}^{\alpha + \mu/2 - 1/2} \over x^{\nu}}\,\dd x
\,\right\vert_{\ \alpha\ =\ 0}
\\[5mm] & \stackrel{x\ \mapsto\ 1/x}{=}\,\,\,
\left.{1 \over 4}\,\partiald{}{\alpha}\int_{1}^{0}
{\pars{1/x - 1}^{\alpha + \mu/2 - 1/2} \over \pars{1/x}^{\nu}}\,
\pars{-\,{\dd x \over x^{2}}}
\right\vert_{\ \alpha\ =\ 0}
\\[5mm] & =
\left.{1 \over 4}\,\partiald{}{\alpha}\int_{0}^{1}
x^{\nu - \alpha - \mu/2 - 3/2}\pars{1 - x}^{\alpha + \mu/2 - 1/2}\,
\dd x\,\right\vert_{\ \alpha\ =\ 0}
\\[5mm] & =
{1 \over 4}\,\partiald{}{\alpha}\bracks{%
\Gamma\pars{\nu - \alpha - \mu/2 - 1/2}\Gamma\pars{\alpha + \mu/2 + 1/2} \over \Gamma\pars{\nu}}_{\ \alpha\ =\ 0}
\\[5mm] & =
\bbx{{\Gamma\pars{\mu/2 + 1/2}\Gamma\pars{\nu - \mu/2 - 1/2} \over
4\Gamma\pars{\nu}}
\bracks{H_{\mu/2 - 1/2} - H_{\nu - \mu/2 - 3/2}}}
\end{align}

$\ds{H_{z}}$ is a Harmonic Number.


$$
\begin{array}{|c|c|}\hline
\ds{\mu \setminus \nu} & \ds{I_{\mu\nu}}
\\ \hline
\ds{0 \setminus 2} & \ds{-\,{\pi \over 4}}
\\ \hline
\ds{0 \setminus 3} & \ds{-\,{\pi \over 4}}
\\ \hline
\end{array}
$$


Note that $\ds{\bbx{I_{\mu,\mu + 1} = 0}}$ because, in such a case,
  $\ds{H_{\mu/2 - 1/2} = H_{\nu - \mu/2 - 3/2}}$.

A: For the first part of the question we can substitute  $x=\frac{1}{t}$ in order to get: $$I=\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x=\int^0_\infty \frac{\ln\left(\frac{1}{t}\right)}{t^2\left(1+\frac{1}{t^2}\right)^3}\frac{-dt}{t^2}=\int_0^\infty \frac{t^2 \ln\left(\frac{1}{t}\right)}{(1+t^2)^3}dt=-I$$
So we just saw that $I=-I\Rightarrow I=0$


I am currently really interested in proving this
  $$ \int_0^\infty \frac{\ln(x)}{(1+x^2)^2} {\rm d}x = \int_0^\infty \frac{\ln(x)}{(1+x^2)^3} {\rm d}x = -\frac{\pi}{4} $$

Well now that we showed that both integrals are equal it's enough to compute only one of it. 
$$\Omega=\int_0^\infty \frac{\ln x}{(1+x^2)^2}dx\overset{x=\tan t}=\int_0^\frac{\pi}{2} \ln(\tan t)\cos^2 t\,\mathrm dt =\frac12\int_0^\frac{\pi}{2} \ln(\tan t) (1+\cos(2t))dt$$
Now we will split into two integrals and show that the first one vanishes using the following property of the definite integrals:
$$\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$$
$$J=\int_0^\frac{\pi}{2}\ln(\tan t)dt=\int_0^\frac{\pi}{2} \ln(\cot t)dt=-\int_0^\frac{\pi}{2} \ln (\tan t)dt=-J\Rightarrow J=0$$
$$\Rightarrow \Omega=\frac12 \int_0^\frac{\pi}{2}\ln(\tan t)\cos (2t)\mathrm dt=\frac12 \int_0^\frac{\pi}{2} \ln(\tan t)\left(\frac12 \sin(2t)\right)' \mathrm dt=$$
$$=\frac14 \underbrace{\ln(\tan t)\sin(2t)\bigg|_0^\frac{\pi}{2}}_{=0}-\frac14\int_0^\frac{\pi}{2} \frac{\sec^2 t}{\tan t}\sin(2t)\mathrm dt=-\frac12 \int_0^\frac{\pi}{2} dt=-\frac{\pi}{4}$$
A: $$\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x=
\int_0^1 \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x+
\int_1^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x
$$
Then change in first integral $x=\frac{1}{t}$.
