How to evaluate the following integral? $\int \ln(e^x + c)~\mathrm dx$ I can't seem to find an answer for this kind of integration, and I'd like to know if there is an answer for it, and if yes what is it.
$$\int \ln(e^x + {c})~\mathrm dx\,,$$
where $c$ is a constant.
My teacher keeps avoiding me after I asked him, so I appreciate any help because I've tried everything I know.

Another approach is
$$
\begin{align}
\int\log(e^x+c)\,\mathrm{d}x
&=\frac{x^2}{2}+\int\log(1+ce^{-x})\,\mathrm{d}x\\
&=\frac{x^2}{2}+\int\left(ce^{-x}-\frac{c^2}{2}e^{-2x}+\frac{c^3}{3}e^{-3x}-\dots\right)\,\mathrm{d}x\\
&=\frac{x^2}{2}+\left(-ce^{-x}+\frac{c^2}{2^2}e^{-2x}-\frac{c^3}{3^2}e^{-3x}+\dots\right)+C\\
&=\frac{x^2}{2}+\mathrm{Li}_2\left(-ce^{-x}\right)+C
\end{align}
$$
which should be the same as Random Variable's answer using the Inversion Formula proven in this answer:
$$
\mathrm{Li}_2(x)=-\frac{\pi^2}{6}-\mathrm{Li}_2(1/x)-\frac12\log(-x)^2
$$
