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!
 A: Hint:
$$\ln a-\ln b=\ln\frac ab.$$
A: Hint:
$$\frac { 1 }{ 1+{ e }^{ t } } =\frac { 1 }{ { e }^{ t }\left( 1+{ e }^{ -t } \right)  } =\frac { { e }^{ -t } }{ 1+{ e }^{ -t } } \\ \int { \frac { dt }{ 1+{ e }^{ t } }  } =-\int { \frac { d\left( 1+{ e }^{ -t } \right)  }{ 1+{ e }^{ -t } }  } $$
A: 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 $1+e^{-t}=u\implies e^{-t}dt=-du$
$$=\int_{1+e^{-x}}^{1}\frac{(-du)}{u}$$
$$=-\int_{1+e^{-x}}^{1}\frac{du}{u}$$
$$=\int_{1}^{1+e^{-x}}\frac{du}{u}$$
$$=\left[\ln|u|\right]_{1}^{1+e^{-x}}$$
$$=\ln|1+e^{-x}|-\ln|1|$$
$$=\ln\left(1+e^{-x}\right)$$
$$=\ln\left(1+\frac{1}{e^{x}}\right)$$
$$=\ln\left(\frac{1+e^x}{e^{x}}\right)$$
$$=\ln\left(1+e^x\right)-\ln\left({e^{x}}\right)$$
$$=\color{red}{\ln(1+e^x)-x}$$
A: 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:
$$
\int_x^\infty \dfrac{1}{1+e^t} dt=f(x)=\log(e^x+1)-x
$$
A: The indefinite integral after you undo your substitution is $\displaystyle \ln \frac{e^t}{1+e^t} + C$.
The lower bound is easy, the value of the integral here is simply $\displaystyle \ln \frac{e^x}{1+e^x}$.
The value of integral at the upper bound is:
$$\lim_{x \to \infty} \ln \frac{e^x}{1+e^x}$$
and that's the natural log of an indeterminate form ($\frac{\infty}{\infty}$)
If you let $f(x) = e^x$ and $g(x) = 1+e^x$, you can use L' Hopital's rule to find that:
$$\lim_{x \to \infty} \frac{f(x)}{g(x)} = \lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \lim_{x \to \infty} \frac{e^x}{e^x} = 1$$
So the value of the integral at the upper bound is $\ln 1 = 0$.
And so the definite integral is: $\displaystyle 0 - \ln \frac{e^x}{1+e^x} = \ln(1 + e^{-x})$.
