Answer
$$\pi $$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^\infty {\frac{4}{{16 + {x^2}}}dx} \cr
& {\text{By Definition }}\int_{ - \infty }^\infty {f\left( x \right)} dx = \int_{ - \infty }^c {f\left( x \right)} dx + \int_c^\infty {f\left( x \right)} dx \cr
& \int_{ - \infty }^\infty {\frac{4}{{16 + {x^2}}}dx} = \int_{ - \infty }^0 {\frac{4}{{16 + {x^2}}}dx} + \int_0^\infty {\frac{4}{{16 + {x^2}}}dx} \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ so}} \cr
& and \cr
& \int_{ - \infty }^b {f\left( x \right)dx = \mathop {\lim }\limits_{a \to - \infty } } \int_a^b {f\left( x \right)} dx,{\text{ so}} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\frac{4}{{16 + {x^2}}}dx} + \mathop {\lim }\limits_{b \to \infty } \int_0^b {\frac{4}{{16 + {x^2}}}dx} \cr
& {\text{Integrating}} \cr
& = 4\mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{1}{4}{{\tan }^{ - 1}}\left( {\frac{x}{4}} \right)} \right]_a^0 + 4\mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{4}{{\tan }^{ - 1}}\left( {\frac{x}{4}} \right)} \right]_0^b \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left[ {{{\tan }^{ - 1}}\left( {\frac{x}{4}} \right)} \right]_a^0 + \mathop {\lim }\limits_{b \to \infty } \left[ {{{\tan }^{ - 1}}\left( {\frac{x}{4}} \right)} \right]_0^b \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left[ {{{\tan }^{ - 1}}\left( {\frac{0}{4}} \right) - {{\tan }^{ - 1}}\left( {\frac{a}{4}} \right)} \right] + \mathop {\lim }\limits_{b \to \infty } \left[ {{{\tan }^{ - 1}}\left( {\frac{b}{4}} \right) - {{\tan }^{ - 1}}\left( {\frac{0}{4}} \right)} \right] \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left[ { - {{\tan }^{ - 1}}\left( {\frac{a}{4}} \right)} \right] + \mathop {\lim }\limits_{b \to \infty } \left[ {{{\tan }^{ - 1}}\left( {\frac{b}{4}} \right)} \right] \cr
& {\text{Evaluate the limits}} \cr
& = - {\tan ^{ - 1}}\left( {\frac{{ - \infty }}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{\infty }{4}} \right) \cr
& = - \left( { - \frac{\pi }{2}} \right) + \left( {\frac{\pi }{2}} \right) \cr
& = \pi \cr} $$
