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.

Partial answer.
Assuming $s$ is an even positive integer,
$\begin{align}G(s)&=\int_{0}^{\infty}\frac{\ln^{s}(t)\ln(1+t)}{t(1+t^2)}\mathrm dt\\
&=\int_{0}^{1}\frac{\ln^{s}(t)\ln(1+t)}{t(1+t^2)}\mathrm dt+\int_{1}^{\infty}\frac{\ln^{s}(t)\ln(1+t)}{t(1+t^2)}\mathrm dt\end{align}$
In the latter integral perform the change of variable $y=\dfrac{1}{t}$,
$\begin{align}G(s)&=\int_{0}^{1}\frac{\ln^{s}(t)\ln(1+t)}{t(1+t^2)}\mathrm dt+\int_{0}^{1}\frac{(-1)^st\ln^{s}(t)\ln\left(\frac{1+t}{t}\right)}{1+t^2}\mathrm dt\\
&=\int_{0}^{1}\frac{\ln^{s}(t)\ln(1+t)}{t}\mathrm dt-\int_{0}^{1}\frac{t\ln^{s}(t)\ln(1+t)}{1+t^2}\mathrm dt+\int_{0}^{1}\frac{(-1)^st\ln^{s}(t)\ln\left(1+t\right)}{1+t^2}\mathrm dt+\\
&\int_{0}^{1}\frac{(-1)^{s+1}t\ln^{s+1}(t)}{1+t^2}\mathrm dt\\
&=\int_{0}^{1}\frac{\ln^{s}(t)\ln(1+t)}{t}\mathrm dt+\int_{0}^{1}\frac{(-1)^{s+1}t\ln^{s+1}(t)}{1+t^2}\mathrm dt
\end{align}$
In the latter integral perform the change of variable $\displaystyle y=t^2$,
$\begin{align}G(s)&=\int_{0}^{1}\frac{\ln^{s}(t)\ln(1+t)}{t}\mathrm dt+\frac{(-1)^{s+1}}{2^{s+2}}\int_{0}^{1}\frac{\ln^{s+1}(t)}{1+t}\mathrm dt\\
&=\int_{0}^{1}\frac{\ln^{s}(t)\ln(1+t)}{t}\mathrm dt-\frac{1}{2^{s+2}}\int_{0}^{1}\frac{\ln^{s+1}(t)}{1+t}\mathrm dt\\
\end{align}$
Perform integration by parts,
$\begin{align}H(s)&=\int_0^1\frac{\ln(1+t)\ln^{s} t}{t}\mathrm dt\\
&=\Big[\ln t \ln^s t\ln(1+t)\Big]_0^1-\int_0^1 \frac{\ln^{s+1} t}{1+t}\mathrm dt-s\int_0^1 \frac{\ln(1+t)\ln^{s} t}{t}\mathrm dt\\
&=-\int_0^1 \frac{\ln^{s+1} t}{1+t}\mathrm dt-sH(s)
\end{align}$
Therefore,
$\begin{align}H(s)=-\frac{1}{s+1}\int_0^1 \frac{\ln^{s+1} t}{1+t}\mathrm dt\end{align}$
$\begin{align}G(s)&=-\left(\frac{1}{2^{s+2}}+\frac{1}{s+1}\right)\int_0^1 \frac{\ln^{s+1} t}{1+t}\mathrm dt\end{align}$
On the other hand,
$\begin{align}\int_0^1 \frac{\ln^{s+1} t}{1+t}\mathrm dt&=\int_0^1 \frac{\ln^{s+1} t}{1-t}\mathrm dt-\int_0^1 \frac{2t\ln^{s+1} t}{1-t^2}\mathrm dt\end{align}$
In the latter integral perform the change of variable $y=x^2$,
$\begin{align}\int_0^1 \frac{\ln^{s+1} t}{1+t}\mathrm dt&=\int_0^1 \frac{\ln^{s+1} t}{1-t}\mathrm dt-\frac{1}{2^{s+1}}\int_0^1 \frac{\ln^{s+1} t}{1-t}\mathrm dt\\
&=\left(1-\frac{1}{2^{s+1}}\right)\int_0^1 \frac{\ln^{s+1} t}{1-t}\mathrm dt\\
&=(-1)^{s+1}\left(1-\frac{1}{2^{s+1}}\right)(s+1)!\zeta(s+2)\\
&=-\left(1-\frac{1}{2^{s+1}}\right)(s+1)!\zeta(s+2)
\end{align}$
Therefore,
If $s$ is an even positive integer,
$\begin{align}\boxed{G(s)=\left(\frac{1}{2^{s+2}}+\frac{1}{s+1}\right)\left(1-\frac{1}{2^{s+1}}\right)(s+1)!\zeta(s+2)}\end{align}$
