How do I evaluate $\int \frac{\mathrm{d}x}{e^x + 1} $? How do I solve evaluate
$$\int \frac{\mathrm{d}x}{e^x + 1}\ ?$$
I know that I have to use $u$ substitution but I can't seem to find something to substitute with.
 A: It also equals $$1-\frac{e^x}{1+e^x}$$
A: So:
$$\int\frac{1}{e^x+1}dx=\int\frac{e^{-x}}{1+e^{-x}}dx=-\int \frac{(1+e^{-x})'}{1+e^{-x}}dx=-\ln(1+e^{-x})+C=\ln(\frac{1}{1+e^{-x}})+C$$
A: Another possibility, if you like series solutions:
\begin{align*}
\int \frac{1}{e^x + 1}dx &= \int \sum_{n=0}^\infty (-1)^ne^{nx}dx \\
&=\sum_{n=0}^\infty (-1)^n\int e^{nx}{dx} \\
&= C + x + \sum_{n=1}^\infty (-1)^n \frac{1}{n}e^{nx} \\
&= C + x - \log(1 + e^x)
\end{align*}
A: Using the above hint that $\frac{1}{\mathrm{e}^x+1}=1-\frac{\mathrm{e}^x}{\mathrm{e}^x+1}$, we get
\begin{align*}
\int\frac{1}{\mathrm{e}^x+1}\,\mathrm{d}x&=\int 1-\frac{\mathrm{e}^x}{\mathrm{e}^x+1}\,\mathrm{d}x \\
&= x-\int\frac{\left(\mathrm{e}^x+1\right)'}{\mathrm{e}^x+1}\,\mathrm{d}x \\
&= x-\log\left(\mathrm{e}^x+1\right)+K
\end{align*}
A: Once again I'm late but better late than never. Let me think first what kind of $u$ substitution should I use because I wanna give the OP different substitution.
Aha! Let $u=e^x+1$ then $e^x=u-1\;\Rightarrow\; x=\ln(u-1)\;\Rightarrow\; dx=\dfrac1{u-1}\ du$, and the integral turns out to be
$$
\begin{align}
\int\frac{1}{e^x+1}\ dx&=\int\frac1u\cdot\dfrac1{(u-1)}\ du\\
&=\int\left[\dfrac1{u-1}-\frac1u\right]\ du\\
&=\ln(u-1)-\ln u+C\tag1\\
&=\ln e^x-\ln (e^x+1)+C\\
&=x-\ln (e^x+1)+C\\
\end{align}
$$
or from $(1)$ we also obtain
$$
\begin{align}
\int\frac{1}{e^x+1}\ dx
&=\ln(u-1)-\ln u+C\\
&=\ln\left(\frac{u-1}{u}\right)+C\\
&=\ln\left(\frac{e^x}{e^x+1}\right)+C\\
&=\ln\left(\frac{1}{1+e^{-x}}\right)+C\\
&=-\ln \left(1+e^{-x}\right)+C.
\end{align}
$$
Done! :)
A: If $u=e^x$ then $du=e^x dx$ and $du= u dx$: this in turn implies $dx= du/u$ so that the integrand gets 
$$
\frac{1}
{(u+1)u} = \frac{1}{u} - \frac{1}{u+1}
$$ 
which integrates into $\ln(u) - \ln(u+1)$ which is $\ln(e^x) - \ln(e^x+1)=x-\ln(e^x + 1)$
A: Setting $\displaystyle e^x=u,e^x\ dx=du\iff dx=\frac{du}{e^x}=\frac{du}u$$$\int\frac{dx}{e^x+1}=\int\frac{du}{u(u+1)}$$
Now $\displaystyle\frac1{u(u+1)}=\frac{u+1-u}{u(u+1)}=\frac1u-\frac1{u+1}$
A: HINT:
$$\frac1{e^x+1}=\frac{e^{-x}}{1+e^{-x}}$$
OR
$$\frac1{e^x+1}=1-\frac{e^x}{e^x+1} $$
A: In the interest of making things appear as simple as they are I am posting after a large number of others have posted.
$$
\int \frac{dx}{e^x+1} = \int \frac 1 {e^x(e^x+1)} \Big( e^x\,dx\Big) = \cdots
$$
(And then partial fractions.)
A: Avoiding substitution (as request by another question)
\begin{eqnarray*}
\int \frac{dx}{1+e^x} &=& \int \left( 1-\frac{e^x}{1+e^x} \right) dx \\
&=& \int \left( 1+\sum_{n=1}^{\infty} (-1)^n e^{nx}  \right) dx \\
&=& x+\sum_{n=1}^{\infty} \frac{(-1)^n}{n}  e^{nx}  +C \\
&=& x- \ln(1+  e^x) +C.
\end{eqnarray*}
