real analysis - How to evaluate $int_{-infty}^{infty}frac{xarctanfrac1x log(1+x^2)}{1+x^2}dx$

While browsing similar questions on this site I came up with the following integral because I thought I could evaluate it.
$$I=\int_{-\infty}^{\infty}\frac{x\arctan x\ \log(1+x^2)}{1+x^2}dx$$
I've been able to simplify it a bit. We first notice that
$$I=\int_0^{\infty}\frac{\arctan x\ \log(1+x^2)}{1+x^2}2xdx$$
The substitution $u=x^2+1$ gives
$$I=\int_1^{\infty}\arctan\sqrt{u-1}\ \log u\ \frac{du}u$$
Then $w=\log u$ gives

$$I=\int_{0}^{\infty}\arctan\sqrt{e^w-1}\ dw$$
Which I do not know how to proceed with.

Another approach I tried was this. Starting with the original integral,

$x=\tan u$: $$I=2\int_0^{\pi/2}u\tan u\log\sec^2u\ du$$
Which I also do not know how to do. Please help me proceed or give me a value of the integral (and show how you got it).

If no closed form exists (AKA you have an answer in terms of a series or special function), I'm fine with that.

cheers!

Edit: In the comments it is discussed that the integral is not integrable over the positive reals, but the following related integral is:

$$J=\int_0^{\infty}\frac{\log(1+x^2)\arctan\frac1x}{1+x^2}xdx$$
So. How do we find the value for $J$?

Answer

Integration by parts reduces the original problem to the evaluation of
$$ \int_{0}^{+\infty}\frac{\log^2(1+x^2)}{1+x^2}\,dx $$
which is pretty straightforward: since

$$ \int_{0}^{+\infty}(1+x^2)^{s-1}\,dx =\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}-s\right)}{\Gamma(1-s)}$$
by applying $\frac{d^2}{ds^2}$ to both sides, then considering $\lim_{s\to 0^+}$, we get:
$$ \int_{0}^{+\infty}\frac{\log^2(1+x^2)}{1+x^2}\,dx =\frac{\pi^3}{6}+2\pi\log^2(2).$$
You may find another example of this technique at page 81 of my notes.