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!

Notice, $$\int_{x}^{\infty}\frac{dt}{1+e^t}$$
$$=\int_{x}^{\infty}\frac{dt}{e^t(e^{-t}+1)}$$
$$=\int_{x}^{\infty}\frac{e^{-t} dt}{1+e^{-t}}$$
Let $1+e^{-t}=u\implies e^{-t}dt=-du$
$$=\int_{1+e^{-x}}^{1}\frac{(-du)}{u}$$
$$=-\int_{1+e^{-x}}^{1}\frac{du}{u}$$
$$=\int_{1}^{1+e^{-x}}\frac{du}{u}$$
$$=\left[\ln|u|\right]_{1}^{1+e^{-x}}$$
$$=\ln|1+e^{-x}|-\ln|1|$$
$$=\ln\left(1+e^{-x}\right)$$
$$=\ln\left(1+\frac{1}{e^{x}}\right)$$
$$=\ln\left(\frac{1+e^x}{e^{x}}\right)$$
$$=\ln\left(1+e^x\right)-\ln\left({e^{x}}\right)$$
$$=\color{red}{\ln(1+e^x)-x}$$
