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

Sep 9, 2016

$\frac{- 2 \left({e}^{x} + 2\right) \sqrt{1 - {e}^{x}}}{3} + C$

#### Explanation:

We have:

$I = \int {e}^{2 x} / \sqrt{1 - {e}^{x}} \mathrm{dx}$

Let $u = {e}^{x}$. This implies that $\mathrm{du} = {e}^{x} \mathrm{dx}$. We can write ${e}^{2 x}$ as ${e}^{x} \left({e}^{x}\right)$:

$I = \int \frac{{e}^{x} \left({e}^{x}\right) \mathrm{dx}}{\sqrt{1 - {e}^{x}}} = \int \frac{u}{\sqrt{1 - u}} \mathrm{du}$

Letting $v = 1 - u$, such that $\mathrm{dv} = - \mathrm{du}$, and manipulating to show that $u = 1 - v$:

$I = - \int \frac{1 - v}{\sqrt{v}} \mathrm{dv} = \int \left({v}^{\frac{1}{2}} - {v}^{- \frac{1}{2}}\right) \mathrm{dv}$

Integrating using the $\int {v}^{n} \mathrm{dv} = {v}^{n + 1} / \left(n + 1\right) , n \ne - 1$ rule:

$I = {v}^{\frac{3}{2}} / \left(\frac{3}{2}\right) - {v}^{\frac{1}{2}} / \left(\frac{1}{2}\right) = \frac{2}{3} {v}^{\frac{3}{2}} - 2 {v}^{\frac{1}{2}} = \frac{2 \sqrt{v} \left(v - 3\right)}{3}$

Since $v = 1 - u$, and $u = {e}^{x}$, so $v = 1 - {e}^{x}$:

$I = \frac{2 \sqrt{1 - {e}^{x}} \left(1 - {e}^{x} - 3\right)}{3} = \frac{- 2 \left({e}^{x} + 2\right) \sqrt{1 - {e}^{x}}}{3} + C$