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?

Hint do another substitution. It is convenient to substitute only $e^u$ and not all the denominator.
$$\int\dfrac{1}{1+e^{-x}}\, dx = -\int\dfrac{1}{1+e^{u}}\, du$$
Now substitute $s=e^u, \dfrac{ds}{e^u}= du$ and use partial fraction decomposition
$$\dfrac{1}{s(s+1)} =-\left( \dfrac{1}{s}-\dfrac{1}{s-1}\right)$$
