Answer
$$2$$
Work Step by Step
Let us consider that $I=\int_{0}^{\infty} \dfrac{e^{-x}}{\sqrt {1-e^{-x}}} dx$
Integrate the integral by using limits.
$I=\lim\limits_{z \to 0^{+}}\int_{z}^{1} \dfrac{e^{-x}}{\sqrt {1-e^{-x}}} dx +\lim\limits_{z \to \infty}\int_{0}^{z} \dfrac{e^{-x}}{\sqrt {1-e^{-x}}} dx \\=\lim\limits_{z \to 0^{+}}[2 \sqrt {1-e^{-x}}]_z^1+\lim\limits_{z \to \infty}[2 \sqrt {1-e^{-x}}]_0^z\\= \lim\limits_{z \to 0^{+}}[2 \sqrt {1-e^{-1}}-2 \sqrt {1-e^{-a}}]+\lim\limits_{z \to \infty}[2 \sqrt {1-e^{-b}}-2 \sqrt {1-e^{-1}}]\\=2-0\\=2$
