Answer
$$\frac{\pi }{4}$$
Work Step by Step
$$\eqalign{
& \int_0^\infty {\frac{1}{{{e^x} + {e^{ - x}}}}dx} \cr
& {\text{Multiply the numerator and denominator by }}{e^x} \cr
& = \int_0^\infty {\frac{{{e^x}}}{{{e^{2x}} + 1}}dx} \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_0^\infty {\frac{{{e^x}}}{{{e^{2x}} + 1}}dx} = \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{{{e^x}}}{{{e^{2x}} + 1}}dx} \cr
& {\text{Integrating}} \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {{{\tan }^{ - 1}}\left( {{e^x}} \right)} \right]_0^b \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {{{\tan }^{ - 1}}\left( {{e^b}} \right) - {{\tan }^{ - 1}}\left( {{e^0}} \right)} \right] \cr
& = \mathop {\lim }\limits_{b \to \infty } \left[ {{{\tan }^{ - 1}}\left( {{e^b}} \right) - \frac{\pi }{4}} \right] \cr
& {\text{Evaluate the limit}} \cr
& = {\tan ^{ - 1}}\left( {{e^\infty }} \right) - \frac{\pi }{4} \cr
& = \frac{\pi }{2} - \frac{\pi }{4} \cr
& = \frac{\pi }{4} \cr} $$
