# Find the integral below?

## $\setminus \int \frac{{e}^{2} x}{1 + {e}^{4 x}} \mathrm{dx}$

May 13, 2018

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

#### Explanation:

Rewrite in terms of powers, recalling that ${e}^{a x} = {\left({e}^{x}\right)}^{a}$:

$\int {\left({e}^{x}\right)}^{2} / \left(1 + {\left({e}^{x}\right)}^{4}\right) \mathrm{dx}$

Let $u = {e}^{x}$

Then, $\mathrm{du} = {e}^{x} \mathrm{dx}$.

We see that since ${\left({e}^{x}\right)}^{2} = {e}^{x} \cdot {e}^{x} , \mathrm{du}$ shows up in our integral:

$\int \frac{{e}^{x} \cdot \textcolor{red}{{e}^{x}}}{1 + {\left({e}^{x}\right)}^{4}} \textcolor{red}{\mathrm{dx}}$

While we can write the remaining ${e}^{x}$ in the numerator as $u ,$ since we originally said $u = {e}^{x}$.

$\int \frac{u}{1 + {u}^{4}} \mathrm{du} = \int \frac{u}{1 + {\left({u}^{2}\right)}^{2}} \mathrm{du}$

We can make a second substitution here.

Let $v = {u}^{2}$

Then, $\mathrm{dv} = 2 u \mathrm{du} \to \frac{1}{2} \mathrm{dv} = u \mathrm{du}$, as $u \mathrm{du}$ shows up in the numerator of the integral

$\int \frac{\textcolor{red}{u}}{1 + {\left({u}^{2}\right)}^{2}} \textcolor{red}{\mathrm{du}}$

Factoring the $\frac{1}{2}$ out of the integral, we get the standard integral

$\frac{1}{2} \int \frac{\mathrm{dv}}{1 + {v}^{2}} = \frac{1}{2} \arctan \left(v\right) + C$

Rewrite in terms of $u :$

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

Rewrite in terms of $x :$

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

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