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.

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:
$$\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$$
$$J=\int_0^\frac{\pi}{2}\ln(\tan t)dt=\int_0^\frac{\pi}{2} \ln(\cot t)dt=-\int_0^\frac{\pi}{2} \ln (\tan t)dt=-J\Rightarrow J=0$$
$$\Rightarrow \Omega=\frac12 \int_0^\frac{\pi}{2}\ln(\tan t)\cos (2t)\mathrm dt=\frac12 \int_0^\frac{\pi}{2} \ln(\tan t)\left(\frac12 \sin(2t)\right)' \mathrm dt=$$
$$=\frac14 \underbrace{\ln(\tan t)\sin(2t)\bigg|_0^\frac{\pi}{2}}_{=0}-\frac14\int_0^\frac{\pi}{2} \frac{\sec^2 t}{\tan t}\sin(2t)\mathrm dt=-\frac12 \int_0^\frac{\pi}{2} dt=-\frac{\pi}{4}$$
