Integration and evaluation I have to evaluate this integral here 
$$\int_x^\infty \frac{1}{1+e^t} dt$$
I've tried with the substitution $e^t=k$, but at the end I get $\log|k|-\log|1+k| $, which evaluated between the extrema doesn't converge... I hope that somebody can help!

Using the substitution $e^t=u$ we have:
$$
\int \dfrac{1}{1+e^t} dt=\int \dfrac{1}{u(1+u)} du
$$
that, using partial fraction, become
$$
\int \left(\dfrac{1}{u}-\dfrac{1}{1+u} \right) du = \log u -\log(1+u) +c =
$$
$$
= \log \left( \dfrac{e^t}{1+e^t} \right) +c
$$
Now taking the limit for $ t\rightarrow \infty$ we have:
$$
\lim_{t\rightarrow \infty}\left[ \log \left( \dfrac{e^t}{1+e^t} \right)\right]=
\lim_{t\rightarrow \infty}\left[ \log \left( \dfrac{1}{1+1/e^t} \right) \right]=0
$$
and substituting also the other limit of integration $t=x$ we find:
$$
\int_x^\infty \dfrac{1}{1+e^t} dt=f(x)=\log(e^x+1)-x
$$
