I want to compute the integral $$2\pi\int f(x) \sqrt{1+f'(x)^2} dx$$ where $f(x)=\dfrac{1}{e^x}$.
I used maple and I found that the answer is: $$\pi e^{-2x} \left[e^{2x} \arctan\left(\sqrt{e^{2x}-1}\right) - \sqrt{e^{2x}-1}\right] $$ but I can't find a way to prove it on the paper. Any help would be apreciated.