Trying to evaluate $\int_{0}^{\infty}\frac{\ln^{s}(t)\ln(1+t)}{t(1+t^2)}\mathrm dt$ $$G(s)=\int_{0}^{\infty}\frac{\ln^{s}(t)\ln(1+t)}{t(1+t^2)}\mathrm dt$$
Trying a substitution of $t=e^x$
$$G(s)=\int_{-\infty}^{\infty}\frac{x^s\ln(1+e^x)}{1+e^{2x}}\mathrm dx$$
I trying to evaluate $G(s)$, but unable to. Can anyone help? Thank you.

Assuming $s\in\mathbb{N}$, you may consider that
$$ \int_{0}^{+\infty}\frac{t^{a-1}\log(1+t)}{1+t^2}\,\mathrm dt=\frac{\pi}{4\sin(\pi a)}\left[H_{-1/2-a/4}-H_{-a/4}+2\log(2)\cos\frac{\pi a}{2}+\pi\sin\frac{\pi a}{2}\right] $$
for any $a$ in a neighbourhood of the origin. In order to compute the wanted integral you may simply differentiate (wrt to $a$) the RHS multiple times, then consider the limit as $a\to 0$.
This should work also by considering fractional derivatives, if defined in terms of the Laplace (inverse) transform. Ultimately we are just dealing with $D^{\alpha}\log\Gamma(z)$.
