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?

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