Improper Integral Question $ \int^{\infty}_0 \frac{\mathrm dx}{1+e^{2x}}$ I want to check if it's improper integral or not 
$$ \int^{\infty}_0 \frac{\mathrm dx}{1+e^{2x}}.$$
What I did so far is : 
set $t=e^{x} \rightarrow \mathrm dt=e^x\mathrm dx \rightarrow \frac{\mathrm dt}{t}=dx
$ so the new integral is:
 $$ \int^{\infty}_0 \frac{\mathrm dt}{t(1+t^2)} =  \int^{\infty}_0 \frac{\mathrm dt}{t}-\frac{\mathrm dt}{1+t^{2}}$$
now how I calculate the improper integral, I need to right the $F(x)$ of this integral and then to check the limit?
Thanks!
 A: You made a couple of mistakes. Firstly, you forgot to change the limits of the integration, so your integral is actually $\displaystyle\int_1^\infty \frac{\mathrm{d}t}{t(1+t^2)}$. Furthermore, $\frac{1}{t(1+t^2)} \neq \frac{1}{t}-\frac{1}{1+t^2}$. Rather $\frac{1}{t(1+t^2)} = \frac{1}{t}-\frac{t}{1+t^2}$.
Hence your integral becomes $\displaystyle\int_1^\infty \frac{1}{t}-\frac{t}{1+t^2}\,\mathrm{d}t = \left[\log(t)-\frac{1}{2}\log(1+t^2)\right]_1^\infty = \left[\frac{1}{2}\log\left(\frac{t^2}{1+t^2}\right)\right]_1^\infty$
$$ = \frac{1}{2}\left[\log(1)-\log\left(\frac{1}{2}\right)\right] = \frac{1}{2}\log(2).$$
A: Hints:
First, it must be
$$\frac1{t(1+t^2)}=\frac1t-\frac t{1+t^2}\;\;,\;\;\text{then}:$$
$$\int\limits_1^\infty\left(\frac1 t-\frac t{1+t^2}\right)dt:=\lim_{b\to\infty}\left(\log\frac b{\sqrt{1+b^2}}+\log\sqrt 2\right)$$
A: Does this solution make sense too?
Let   $\displaystyle t=1+e^{2x},$   $x\in(0,\infty), t\in(2,\infty)$
So    $\displaystyle dx = \frac{1}{2(t-1)}dt$,
Then, 
      $\displaystyle\int^{\infty}_0\frac{1}{1+e^{2x}}dx$ 
  $\displaystyle= \frac{1}{2}\int^{\infty}_2\frac{1}{t(t-1)}dt$ 
  (Please pay attention to the changing interval) 
  $\displaystyle= \frac{1}{2}\int^{\infty}_2(\frac{1}{t-1} - \frac{1}{t})dt$ 
  $\displaystyle= \frac{1}{2}\left(\left[\ln{\left(t-1\right)}\right]_2^{\infty} - \left[\ln{t}\right]_2^{\infty}\right)$ 
  $\displaystyle= \frac{1}{2}\lim_{t\to\infty}\ln{(t-1)} - \frac{1}{2}\lim_{t\to\infty}\ln{t} + \frac{1}{2}\ln{2}$
  $\displaystyle= \frac{1}{2}\lim_{t\to\infty}\ln{\frac{t-1}{t}} + \frac{1}{2}\ln{2}$ 
  $\displaystyle= \frac{1}{2}\ln{2}$
A: $$
\begin{aligned}
\int_{0}^{\infty} \frac{d x}{1+e^{2 x}} &=\int_{0}^{\infty} \frac{e^{-2 x}}{e^{-2 x}+1} d x \\
&=-\frac{1}{2} \int_{0}^{\infty} \frac{d\left(e^{-2 x}+1\right)}{e^{-2 x}+1} \\
&=-\frac{1}{2}\left[\ln \left(e^{-2 x}+1\right)\right]_{0}^{\infty} \\
&=\frac{1}{2} \ln 2
\end{aligned}
$$
