Prove that

\begin{equation}
\int_{-\infty}^{\infty} \frac{\ln(1+e^{ax})}{1+e^{bx}}\, dx = \frac{\pi^{2}}{12} \frac{a^{2} + b^{2}}{a b^{2}}.
\end{equation}

(sos440)

Proof:
Let $I$ denote the integral  . With the substitution $t=e^{bx}$, we have
\begin{align*}
I&= \int_{-\infty}^{\infty} \frac{\ln(1+e^{ax})}{1+e^{bx}}\, dx
 = \frac{1}{b} \int_{0}^{\infty} \frac{\ln(1+t^{\frac{a}{b}})}{t(1+t)}\, dt.
\end{align*}
Then with the substitution $t\mapsto\frac{1}{t}$, it follows that
\begin{align*}
I&= \frac{1}{b} \int_{0}^{\infty} \frac{\ln(1+t^{\frac{a}{b}}) - \frac{a}{b} \ln t}{1+t}\, dt.
\end{align*}
Summing two identities above, we have

\begin{align*}
2I&= \frac{1}{b} \int_{0}^{\infty} \left( \frac{\ln(1 + t^{\frac{a}{b}})}{t} - \frac{a}{b} \frac{\ln t}{1 + t} \right) \, dt \\
&= \lim_{R\to\infty} \frac{1}{b} \int_{0}^{R} \left( \frac{\ln(1 + t^{\frac{a}{b}})}{t} - \frac{a}{b} \frac{\ln t}{1 + t} \right) \, dt.
\end{align*}
Now we focus on the integral inside the limit. Simplifying, we have
\begin{align*}
& \frac{1}{b} \int_{0}^{R} \left( \frac{\ln(1 + t^{\frac{a}{b}})}{t} - \frac{a}{b} \frac{\ln t}{1 + t} \right) \, dt \\
&= \frac{1}{a} \int_{0}^{R^{\frac{a}{b}}} \frac{\ln(1 + t)}{t} \, dt - \frac{a}{b^{2}} \int_{0}^{R} \frac{\ln t}{1 + t} \, dt \\
&= \frac{1}{a} \int_{0}^{R^{\frac{a}{b}}} \frac{\ln(1 + t)}{t} \, dt - \frac{a}{b^{2}} \ln R \ln (R+1) + \frac{a}{b^{2}} \int_{0}^{R} \frac{\ln (1+t)}{t} \, dt \\
&= -\frac{1}{a} \mathrm{Li}_{2}(-R^{\frac{a}{b}}) - \frac{a}{b^{2}} \mathrm{Li}_{2} (-R) - \frac{a}{b^{2}} \ln R \ln (R+1).
\end{align*}
But we know that the dilogarithm satisfies
\begin{align*}
-\mathrm{Li}_{2}(-x)
&= \mathrm{Li}_{2}\left( \frac{x}{1+x} \right) + \frac{1}{2}\ln^{2}(x + 1)
= \zeta(2) + \frac{1}{2}\ln^2 x + o(1)
\end{align*}
as $x\to\infty$ . Plugging this identity above, it follows that
\begin{align*}
\frac{1}{b} \int_{0}^{R} \left( \frac{\ln(1 + t^{\frac{a}{b}})}{t} - \frac{a}{b} \frac{\ln t}{1 + t} \right) \, dt
&= \frac{a^2 + b^2}{a b^2} \zeta(2) + o(1)
\end{align*}
as $R\to\infty$, and therefore the proof is complete.