Answer
$\dfrac{\pi}{2}$
Work Step by Step
Let us consider that $I=\int_{0}^{\infty} \dfrac{e^{-x}}{\sqrt {1-e^{-2x}}} dx$
Plug in $a=e^{-x} \implies da=-e^{-x} \ dx$
Integrate the integral by using limits.
$I=\int_{0}^{\infty} \dfrac{e^{-x}}{\sqrt {1-e^{-2x}}} dx \\=\int_{0}^{1} \dfrac{dx}{\sqrt {1-a^2}} dx\\=\lim\limits_{z \to 1}|\sin^{-1} (a)|_0^1\\=\dfrac{\pi}{2}$
