Ways to prove that $\int_0^\infty \frac{x^2 \ln(x)}{(1+x^2)^3} {\rm d}x = 0$.
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: $$\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$$ $$J=\int_0^\frac{\pi}{2}\ln(\tan t)dt=\int_0^\frac{\pi}{2} \ln(\cot t)dt=-\int_0^\frac{\pi}{2} \ln (\tan t)dt=-J\Rightarrow J=0$$ $$\Rightarrow \Omega=\frac12 \int_0^\frac{\pi}{2}\ln(\tan t)\cos (2t)\mathrm dt=\frac12 \int_0^\frac{\pi}{2} \ln(\tan t)\left(\frac12 \sin(2t)\right)' \mathrm dt=$$ $$=\frac14 \underbrace{\ln(\tan t)\sin(2t)\bigg|_0^\frac{\pi}{2}}_{=0}-\frac14\int_0^\frac{\pi}{2} \frac{\sec^2 t}{\tan t}\sin(2t)\mathrm dt=-\frac12 \int_0^\frac{\pi}{2} dt=-\frac{\pi}{4}$$
$$\int_{0}^{+\infty}\frac{x^{2+\alpha}}{(1+x^2)^3}\,dx \stackrel{(*)}{=}\frac{\pi(1-\alpha^2)}{16\cos\frac{\pi \alpha}{2}} $$ $(*)$: 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. $$\color{red}{0}=\frac{d}{d\alpha}\left.\int_{0}^{+\infty}\frac{x^{2+\alpha}}{(1+x^2)^3}\,dx\right|_{\alpha=0}\stackrel{\text{DCT}}{=}\int_{0}^{+\infty}\frac{x^{2}\log x}{(1+x^2)^3}\,dx.$$
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 $\alpha = 2$.