Answer
$\dfrac{\pi}{2}$
Work Step by Step
We can rewrite as: $\dfrac{1}{e^{x}+e^{-x}}=\dfrac{e^x}{(e^{x})^2+1}$
$\int_{-\infty}^{\infty} \dfrac{dx}{e^{x}+e^{-x}}=\int_{-\infty}^{0}\dfrac{dx}{e^{x}+e^{-x}} +\int_{0}^{\infty} \dfrac{dx}{e^{x}+e^{-x}} $
and $\int_{-\infty}^{\infty} \dfrac{dx}{e^{x}+e^{-x}}=\lim\limits_{a \to -\infty} \int_{a}^{0}\dfrac{dx}{e^{x}+e^{-x}} + \lim\limits_{b \to \infty} \int_{0}^{b} \dfrac{dx}{e^{x}+e^{-x}}....(1)$
Let $u =e^x $ and $du =e^x dx$
$\int \dfrac{dx}{e^{x}+e^{-x}}=\int \dfrac{du}{u^2+1}=\tan^{-1} u+C=\tan^{-1} (e^x) +C$
Here, $C$ is an arbitrary Constant.
Equation (1) becomes:
$$\int_{-\infty}^{\infty} \dfrac{dx}{e^{x}+e^{-x}}=\lim\limits_{a \to -\infty} \int_{a}^{0}\dfrac{dx}{e^{x}+e^{-x}} + \lim\limits_{b \to \infty} \int_{0}^{b} \dfrac{dx}{e^{x}+e^{-x}} \\=\lim\limits_{a \to -\infty} [\tan^{-1} (e^x)]_{a}^{0}+\lim\limits_{a \to \infty} [\tan^{-1} (e^x)]_{0}^{b} \\=\lim\limits_{a \to -\infty} [\tan^{-1} (1)-\tan^{-1} e^a]+\lim\limits_{a \to \infty} [\tan^{-1} (e^b)-\tan^{-1} (1)] \\=(\dfrac{\pi}{4}-0)+(\dfrac{\pi}{2}-\dfrac{\pi}{4}) \\=\dfrac{\pi}{2}$$
