Answer
$$\frac{{3\pi }}{8}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^2 {\frac{{dx}}{{{x^2} + 4}}} \cr
& {\text{using the definition 7}}{\text{.8}}{\text{.1 of improper integrals}} \cr
& \,\,\,\int_{ - \infty }^b {f\left( x \right)} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^b {f\left( x \right)} dx \cr
& \cr
& {\text{then}} \cr
& \,\int_{ - \infty }^2 {\frac{{dx}}{{{x^2} + 4}}} = \mathop {\lim }\limits_{a \to - \infty } \int_a^2 {\frac{{dx}}{{{x^2} + 4}}} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \int_a^2 {\frac{{dx}}{{{x^2} + {{\left( 2 \right)}^2}}}} \cr
& \cr
& {\text{Integrate by tables using the formula }}\int {\frac{1}{{{x^2} + {a^2}}}} dx = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + C \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{1}{2}{{\tan }^{ - 1}}\left( {\frac{x}{2}} \right)} \right]_a^2 \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {{{\tan }^{ - 1}}\left( {\frac{2}{2}} \right) - {{\tan }^{ - 1}}\left( {\frac{a}{2}} \right)} \right] \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{\pi }{4} - {{\tan }^{ - 1}}\left( {\frac{a}{2}} \right)} \right] \cr
& \cr
& {\text{calculate the limit when }}a \to - \infty \cr
& = \frac{1}{2}\left[ {\frac{\pi }{4} - {{\tan }^{ - 1}}\left( {\frac{{ - \infty }}{2}} \right)} \right] \cr
& = \frac{1}{2}\left[ {\frac{\pi }{4} - \left( { - \frac{\pi }{2}} \right)} \right] \cr
& = \frac{1}{2}\left( {\frac{{3\pi }}{4}} \right) \cr
& = \frac{{3\pi }}{8} \cr
& {\text{then}}{\text{,}} \cr
& {\text{The integral converges to }}\frac{{3\pi }}{8} \cr} $$
