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 
$$I=\ln (1+e^x) e^x - x- \ln(|1+e^x|) +C $$

For your calculation you made one step (at least) wrong 
$$
\int 1 \to u \neq x
$$
So note to keep the variable you are integrating over. 
To be honest your approach is rather long.
$$
u = 1+\mathrm{e}^x\implies du = \mathrm{e}^xdx
$$
so the integral becomes
$$
\int \ln u \;du = u\ln u - u +c
$$
sub back in for $x$.
