How do I evaluate $\int \frac{\mathrm{d}x}{e^x + 1} $?

Question

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.

Answer 1

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 $\displaystyle\frac1{u(u+1)}=\frac{u+1-u}{u(u+1)}=\frac1u-\frac1{u+1}$

Answer 2

HINT:

$$\frac1{e^x+1}=\frac{e^{-x}}{1+e^{-x}}$$ OR

$$\frac1{e^x+1}=1-\frac{e^x}{e^x+1} $$

Answer 3

It also equals $$1-\frac{e^x}{1+e^x}$$

Answer 4

So:

$$\int\frac{1}{e^x+1}dx=\int\frac{e^{-x}}{1+e^{-x}}dx=-\int \frac{(1+e^{-x})'}{1+e^{-x}}dx=-\ln(1+e^{-x})+C=\ln(\frac{1}{1+e^{-x}})+C$$

Answer 5

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*}

Answer 6

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*}

Answer 7

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! :)

Answer 8

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.)