How do you evaluate the integral int e^t/(e^t+1)dt?

May 6, 2017

$\int \setminus \frac{{e}^{t}}{{e}^{t} + 1} \setminus \mathrm{dt} = \ln \left({e}^{t} + 1\right) + C$

Explanation:

We want to find:

$I = \int \setminus \frac{{e}^{t}}{{e}^{t} + 1} \setminus \mathrm{dt}$

We can perform a simple substitution; Let

$u = {e}^{t} + 1 \implies \frac{\mathrm{du}}{\mathrm{dt}} = {e}^{t}$

If we perform the substitution then we get:

$I = \int \frac{\textcolor{red}{{e}^{t}}}{u} \text{ } \frac{\mathrm{du}}{\textcolor{red}{{e}^{t}}}$
$\setminus \setminus = \int \setminus \frac{1}{u} \setminus \mathrm{du}$
$\setminus \setminus = \ln | u | + C$

And restoring the substitution we get:

$I = \ln \left({e}^{t} + 1\right) + C$