How am I computing $\int e^{x}\ln(1+e^{x})\,dx$ incorrect? I am stuck on this problem.
Problem Evaluate $$\int e^{x} \ln (1+e^{x})$$
Attempt
Integration by Parts:
Let $u=\ln(1+e^x), \ dv = \int e^{x} \ dx$ and we have $\ du = \frac{e^x}{1+e^x}$ and $v=e^x$
$$\int u \ dv= uv - \int v \ du$$
Thus $$I=\ln(1+e^x) e^x - \int e^x \frac{e^x}{1+e^x} \ dx$$
On the second integral apply $u=1+e^x$ and $\ du = e^x \ dx$ and 
$$\begin{split}
I&=\ln (1+e^x) e^x - \int \frac{u-1}{u} \ du \\
&= \ln (1+e^x) e^x - \int 1- \frac{1}{u} \ du
\end{split}$$ which simplifies to 
$$I=\ln (1+e^x) e^x - x- \ln(|1+e^x|) +C $$

You should just let $u=1+e^x$, $du=e^x\,dx$
Then
\begin{eqnarray} \int e^{x}\ln(1+e^{x})dx&=&\int\ln(u)\,du \\
&=&u\ln(u)-u+c
\end{eqnarray}
and take it from there.

