Answer
$$\frac{\pi }{4}$$
Work Step by Step
$$\eqalign{
& \int_{ - \infty }^3 {\frac{{dx}}{{{x^2} + 9}}} \cr
& \cr
& {\text{using the definition 7}}{\text{.8}}{\text{.1 of improper integrals}} \cr
& \,\,\,\int_{ - \infty }^b {f\left( x \right)} dx = \mathop {\lim }\limits_{a \to - \infty } \int_a^b {f\left( x \right)} dx \cr
& \cr
& {\text{then}} \cr
& \,\int_{ - \infty }^3 {\frac{{dx}}{{{x^2} + 9}}} = \mathop {\lim }\limits_{a \to - \infty } \int_a^3 {\frac{{dx}}{{{x^2} + 9}}} \cr
& = \mathop {\lim }\limits_{a \to - \infty } \int_a^3 {\frac{{dx}}{{{x^2} + {{\left( 3 \right)}^2}}}} \cr
& \cr
& {\text{Integrate by tables using the formula }}\int {\frac{1}{{{x^2} + {a^2}}}} dx = \frac{1}{a}{\tan ^{ - 1}}\left( {\frac{x}{a}} \right) + C \cr
& = \mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{x}{3}} \right)} \right]_a^3 \cr
& = \frac{1}{3}\mathop {\lim }\limits_{a \to - \infty } \left[ {{{\tan }^{ - 1}}\left( {\frac{3}{3}} \right) - {{\tan }^{ - 1}}\left( {\frac{a}{3}} \right)} \right] \cr
& = \frac{1}{3}\mathop {\lim }\limits_{a \to - \infty } \left[ {\frac{\pi }{4} - {{\tan }^{ - 1}}\left( {\frac{a}{3}} \right)} \right] \cr
& \cr
& {\text{calculate the limit when }}a \to - \infty \cr
& = \frac{1}{3}\left[ {\frac{\pi }{4} - {{\tan }^{ - 1}}\left( {\frac{{ - \infty }}{3}} \right)} \right] \cr
& = \frac{1}{3}\left[ {\frac{\pi }{4} - \left( { - \frac{\pi }{2}} \right)} \right] \cr
& = \frac{1}{3}\left( {\frac{{3\pi }}{4}} \right) \cr
& = \frac{\pi }{4} \cr
& {\text{then}}{\text{,}} \cr
& {\text{The integral converges to }}\frac{\pi }{4} \cr} $$
