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Evaluate $$\displaystyle\int{\frac{e^{2x}} {\sqrt{1-e^x}}}\ dx.$$

I tried to solve by using integration by parts, but I couldn't find a solution. What method should I use to integrate this?

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  • $\begingroup$ Welcome to Stack Exchange. Can you tell us what you tried to do to solve this problem? $\endgroup$ – Asimov May 19 '14 at 1:39
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    $\begingroup$ Is that supposed to be $e^{2x}$ or $e^2x$, as this drastically changes the difficulty of the integral $\endgroup$ – Triatticus May 19 '14 at 1:45
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Hint: Let $u=1-e^{x}$${}{}{}{}{}{}{}$

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  • $\begingroup$ by substitution ? But this integral its in the "integral by parts list", sorry bad english I tried u = (1-e^x)^-1/2 $\endgroup$ – user151911 May 19 '14 at 1:43
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    $\begingroup$ An alternative way to solve this using integration by parts is to notice that $\frac{e^{2x}}{\sqrt{1-e^{x}}}=\frac{e^{x}}{\sqrt{1-e^{x}}}e^{x}=\bigg(\frac{d}{dx}(2 \sqrt{1-e^{x}})\bigg)e^{x}$ $\endgroup$ – user71352 May 19 '14 at 1:46
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$$ \int\frac{e^x}{\sqrt{1-e^x}}\underbrace{\Big(e^x \, dx\Big)}_{\text{HINT}} $$

After that substitution (the one that is hinted at), use another substitution, namely a rationalizing substitution.

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$$I=\int\frac{e^{2x}}{(1-e^x)^{1/2}}dx$$ $$I=\int\frac{e^x}{(1-e^x)^{1/2}}e^xdx$$ Substitution: $u^2=1-e^x\Rightarrow -2udu=e^xdx$. $$I=\int\frac{1-u^2}{(u^2)^{1/2}}(-2)udu$$ $$I=-2\int(1-u^2)du$$ $$I=2\int(u^2-1)du$$ $$I=2\int u^2du-2\int du$$ $$I=\frac{2u^4}{4}-2u$$ $$I=\frac{(1-e^x)^2}{2}-2\sqrt{1-e^x}+C$$

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