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

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

Solution 2:

HINT:

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

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

Solution 3:

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

Solution 4:

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*}