Answer
$\dfrac{\pi}{2}$
Work Step by Step
We have: $\int_{-\infty}^{\infty} \dfrac{dx}{e^{x}+e^{-x}}=\int_{-\infty}^{0}\dfrac{dx}{e^{x}+e^{-x}} +\int_{0}^{\infty} \dfrac{dx}{e^{x}+e^{-x}}$
or, $=\lim\limits_{a \to -\infty} \int_{a}^{0}\dfrac{dx}{e^{x}+e^{-x}} + \lim\limits_{b \to \infty} \int_{0}^{b} \dfrac{dx}{e^{x}+e^{-x}}$
Re-write as : $\dfrac{1}{e^{x}+e^{-x}}=\dfrac{e^x}{(e^{x})^2+1}$
Consider $u =e^x \implies du =e^x dx$
So, $\int \dfrac{dx}{e^{x}+e^{-x}}=\int \dfrac{du}{u^2+1}=\tan^{-1} u+c=\tan^{-1} (e^x) +c$
and $C$ is an arbitrary constant.
Now, $\int_{-\infty}^{\infty} \dfrac{dx}{e^{x}+e^{-x}}=\lim\limits_{a \to -\infty} \int_{a}^{0}\dfrac{dx}{e^{x}+e^{-x}} + \lim\limits_{b \to \infty} \int_{0}^{b} \dfrac{dx}{e^{x}+e^{-x}}$
or, $=\lim\limits_{a \to -\infty} [\tan^{-1} (e^x)]_{a}^{0}+\lim\limits_{a \to \infty} [\tan^{-1} (e^x)]_{0}^{b}$
or, $=\lim\limits_{a \to -\infty} [\tan^{-1} (1)-\tan^{-1} e^a]+\lim\limits_{a \to \infty} [\tan^{-1} (e^b)-\tan^{-1} (1)]$
or, $=(\dfrac{\pi}{4}-0)+(\dfrac{\pi}{2}-\dfrac{\pi}{4})$
or, $=\dfrac{\pi}{2}$
