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.

$$\int_{0}^{+\infty}\frac{x^{2+\alpha}}{(1+x^2)^3}\,dx \stackrel{(*)}{=}\frac{\pi(1-\alpha^2)}{16\cos\frac{\pi \alpha}{2}} $$
$(*)$: 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.
$$\color{red}{0}=\frac{d}{d\alpha}\left.\int_{0}^{+\infty}\frac{x^{2+\alpha}}{(1+x^2)^3}\,dx\right|_{\alpha=0}\stackrel{\text{DCT}}{=}\int_{0}^{+\infty}\frac{x^{2}\log x}{(1+x^2)^3}\,dx.$$
