integral of $\frac{1}{(1+e^{-x})}$ I make the substitution $u=1+e^{-x}$ which gives $-\dfrac{e^x}{u}\ du$.
Integrating gives me $$-e^x\ln(1+e^{-x}) + C,$$
but the answer is $\ln(e^x +1) + C$.
What am I doing wrong?

Your steps are not quite right. It seems you're skipping a couple of them.
From the substitution $u=1+e^{-x}$, you should get:
$$du=d(1+e^{-x}) = -e^{-x}dx$$
So that:
$$dx = - \frac {du}{e^{-x}}$$
Since $u=1+e^{-x}$ implies $e^{-x}=u-1$, this becomes:
$$dx = - \frac {du}{u-1}$$
So your integral should be:
$$\int \frac {dx} {1+e^{-x}} =\int \frac 1 u \cdot - \frac {du}{u-1} = \int -\frac{du}{u(u-1)}$$
