What is the integral of 1/(e^x + 1)?

integral of 1/(e^x + 1)

The integral of

\frac{1}{e^x + 1}

-\ln \lvert 1 + e^{-x} \rvert + C

.

Solution. We want to determine the integral of

\frac{1}{e^x + 1}

, i.e.:

\begin{align*} \int \frac{1}{e^x + 1} dx = \int \frac{1}{e^x(1 + e^{-x})} dx = \int \frac{e^{-x}}{1 + e^{-x}} dx. \end{align*}

We will apply the substitution method, where

u = 1 + e^{-x}

. Then we know from here that

du = -e^{-x} dx

. So we get the following:

\begin{align*} \int \frac{e^{-x}}{1 - e^{-x}} dx &= \int \frac{-du}{u} \\ &= -\ln \lvert u \rvert + C \\ &= -\ln \lvert 1 + e^{-x} \rvert + C, \end{align*}

where

C

is some constant. So, the integral of

\frac{1}{e^x + 1}

-\ln \lvert 1 + e^{-x} \rvert + C