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

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?

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

## 3 Answers

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

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