# How do you find the antiderivative of (e^(2x)+e^x)/(e^(2x)+1)?

Jul 31, 2016

$= \frac{1}{2} \ln \left({e}^{2 x} + 1\right) + \arctan {e}^{x} + C$

#### Explanation:

$\int \setminus \frac{{e}^{2 x} + {e}^{x}}{{e}^{2 x} + 1} \setminus \mathrm{dx}$

$= \int \setminus \frac{{e}^{2 x}}{{e}^{2 x} + 1} + \frac{{e}^{x}}{{e}^{2 x} + 1} \setminus \mathrm{dx}$

spotting the pattern
$= \int \setminus \frac{d}{\mathrm{dx}} \left(\frac{1}{2} \ln \left({e}^{2 x} + 1\right)\right) + \frac{{e}^{x}}{{e}^{2 x} + 1} \setminus \mathrm{dx}$

$= \frac{1}{2} \ln \left({e}^{2 x} + 1\right) + \int \setminus \frac{{e}^{x}}{{e}^{2 x} + 1} \setminus \mathrm{dx} q \quad \triangle$

for the remaining part we sub $p = {e}^{x} , \mathrm{dp} = {e}^{x} \mathrm{dx} = p \setminus \mathrm{dx}$

so $\int \setminus \frac{{e}^{x}}{{e}^{2 x} + 1} \setminus \mathrm{dx} \to \int \frac{p}{{p}^{2} + 1} \setminus \frac{1}{p} \setminus \mathrm{dp}$

$= \int \frac{1}{{p}^{2} + 1} \setminus \mathrm{dp} = \arctan p$ where $p = {e}^{x}$

so $\triangle$ becomes

$= \frac{1}{2} \ln \left({e}^{2 x} + 1\right) + \arctan {e}^{x} + C$