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.

I'm assuming $c >0$.
Let $u =e^{x}$.
Then $du = e^{x} \ dx = u \ dx$.
And
$$ \begin{align} \int \log(e^{x} + c) \ dx &= \int \frac{\log(u+c)}{u} du \\ &= \int \frac{\log(c(1+\frac{u}{c}))}{u} \ du \\ &= \log (c) \int \frac{1}{u} \ du + \int \frac{\log(1+\frac{u}{c})}{u} du \\ &=\log(c) \log (u) - \text{Li}_{2} \left(-\frac{u}{c} \right) + C  \\ &= \log(c) \log(e^{x}) - \text{Li}_{2} \left(-\frac{e^{x}}{c} \right) + C  \\ &= x \log(c) - \text{Li}_{2} \left(-\frac{e^{x}}{c} \right) + C \end{align}   $$
where $\text{Li}_{2}(x)$ is the dilogarithm function.
EDIT:
Similar to what I did above,
$$ \int \ln(e^{x}-c) \ dx = x\log(c) + \int \frac{\log(\frac{u}{c}-1)}{u} \ du$$
Let $ \displaystyle v= \frac{u}{c}$.
$$ = x \log(c) + \int \frac{\log (v-1)}{v} \ dv$$
Let $w = v-1$.
$$ \begin{align} &= x \log (c) + \int \frac{\log w}{1+w} \ dw \\ &= x \log(c) + \log(w) \log(1+w) - \int \frac{\log(1+w)}{w} \ dw \\ &= x \log(c) + \log(w) \log(1+w) + \text{Li}_{2}(-w) + C \\ &= x \log(c) + \log \left(\frac{e^{x}}{c}-1\right)\log\left(\frac{e^{x}}{c}\right) + \text{Li}_{2} \left( 1-\frac{e^{x}}{c} \right) + C \end{align}$$
This is not in the form given by Wolfram Alpha.  
But using it to evaluate some definite integrals for specific values of $c$, it appears to be correct.
