How do you find the antiderivative of $\frac{e^{2 x}}{\sqrt{1 - e^{x}}}$ $e^(2x)/sqrt(1-e^x)$ ?

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How do you find the antiderivative of $\frac{e^{2 x}}{\sqrt{1 - e^{x}}}$ $e^(2x)/sqrt(1-e^x)$ ?

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Answer 1 · 2016-09-09T21:55:36

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

Explanation:

We have:

$I = \int \frac{e^{2 x}}{\sqrt{1 - e^{x}}} d x$ $I=inte^(2x)/sqrt(1-e^x)dx$

Let $u = e^{x}$ $u=e^x$ . This implies that $d u = e^{x} d x$ $du=e^xdx$ . We can write $e^{2 x}$ $e^(2x)$ as $e^{x} (e^{x})$ $e^x(e^x)$ :

$I = \int \frac{e^{x} (e^{x}) d x}{\sqrt{1 - e^{x}}} = \int \frac{u}{\sqrt{1 - u}} d u$ $I=int(e^x(e^x)dx)/sqrt(1-e^x)=intu/sqrt(1-u)du$

Letting $v = 1 - u$ $v=1-u$ , such that $d v = - d u$ $dv=-du$ , and manipulating to show that $u = 1 - v$ $u=1-v$ :

$I = - \int \frac{1 - v}{\sqrt{v}} d v = \int (v^{\frac{1}{2}} - v^{- \frac{1}{2}}) d v$ $I=-int(1-v)/sqrtvdv=int(v^(1/2)-v^(-1/2))dv$

Integrating using the $\int v^{n} d v = \frac{v^{n + 1}}{n + 1}, n \neq - 1$ $intv^ndv=v^(n+1)/(n+1),n!=-1$ rule:

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

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

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

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How do you find the antiderivative of $\frac{e^{2 x}}{\sqrt{1 - e^{x}}}$ $e^(2x)/sqrt(1-e^x)$ ?

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Explanation:

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