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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: $$I=int_0^infty frac{x^2 ln(x)}{(1+x^2)^3} {rm d}x=int^0_infty frac{lnleft(frac{1}{t}right)}{t^2left(1+frac{1}{t^2}right)^3}frac{-dt}{t^2}=int_0^infty frac{t^2 lnleft(frac{1}{t}right)}{(1+t^2)^3}dt=-I$$
So we just saw that $I=-IRightarrow I=0$

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. Of course it's more rational to take the one with the lower power.
$$Omega=int_0^infty frac{ln x}{(1+x^2)^2}dxoverset{x=tan t}=int_0^frac{pi}{2} ln(tan t)cos^2 t,mathrm dt =frac12int_0^frac{pi}{2} ln(tan t) (1+cos(2t))dt$$
Now we will split into two integrals and first we will show that the first one vanishes. I will do it directly using the following property of the definite integrals:
$$int_a^b f(x)=int_a^b f(a+b-x)dx$$
$$J=int_0^frac{pi}{2}ln(tan t)dt=int_0^frac{pi}{2} ln(cot t)dt=-int_0^frac{pi}{2} ln (tan t)dt=-JRightarrow J=0$$
$$Rightarrow Omega=frac12 int_0^frac{pi}{2}ln(tan t)cos (2t)mathrm dt=frac12 int_0^frac{pi}{2} ln(tan t)left(frac12 sin(2t)right)' mathrm dt=$$
$$=frac14 underbrace{ln(tan t)sin(2t)bigg|_0^frac{pi}{2}}_{=0}-frac14int_0^frac{pi}{2} frac{sec^2 t}{tan t}sin(2t)mathrm dt=-frac12 int_0^frac{pi}{2} dt=-frac{pi}{4}$$

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

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
$$
int_0^infty frac{ln(x)}{x^2 +alpha^2} {rm d}x = frac{pi}{2alpha} ln(alpha)
$$
and use differentiation under the integral sign.

$$
int_0^infty frac{-2alphaln(x)}{(x^2 +alpha^2)^2} {rm d}x
= frac{pi}{2} frac{1-ln(alpha)}{alpha^2} \
implies int_0^infty frac{ln(x)}{(x^2 +alpha^2)^2} {rm d}x
= frac{pi}{4alpha^3} (ln(alpha)-1)
$$

And again, to get
$$
int_0^infty frac{ln(x)}{(x^2 +alpha^2)^3} {rm d}x
= frac{pi}{16alpha^5}(3ln(alpha) - 4)
$$

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 $u=frac1x$)
$$ int_0^infty frac{1}{(1+x^2)(1+x^alpha)} {rm d}x = fracpi4 $$

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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With $ds{Repars{mu} > - 1}$ and
$ds{Repars{nu} > 0}$:

begin{align}
I_{munu} & equiv
bbox[10px,#ffd]{int_{0}^{infty}{x^{mu}lnpars{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}lnpars{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
,rightvert_{ 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
,rightvert_{ 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}}}
rightvert_{ 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,rightvert_{ alpha = 0}
\[5mm] & =
{1 over 4},partiald{}{alpha}bracks{%
Gammapars{nu - alpha - mu/2 - 1/2}Gammapars{alpha + mu/2 + 1/2} over Gammapars{nu}}_{ alpha = 0}
\[5mm] & =
bbx{{Gammapars{mu/2 + 1/2}Gammapars{nu - mu/2 - 1/2} over
4Gammapars{nu}}
bracks{H_{mu/2 - 1/2} - H_{nu - mu/2 - 3/2}}}
end{align}

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

$$int_0^infty frac{x^2 ln(x)}{(1+x^2)^3} {rm d}x=
int_0^1 frac{x^2 ln(x)}{(1+x^2)^3} {rm d}x+
int_1^infty frac{x^2 ln(x)}{(1+x^2)^3} {rm d}x
$$
Then change in first integral $x=frac{1}{t}$.