<s>[INST] 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.
 [/INST] 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 $\alpha = 2$.
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