Answer
$$\frac{\pi }{6}$$
Work Step by Step
$$\eqalign{
& \int_3^\infty {\frac{1}{{x\sqrt {{x^2} - 9} }}} dx \cr
& \frac{1}{{x\sqrt {{x^2} - 9} }}{\text{ has an infinite discontinuity at }}x = 3,{\text{so we can write}} \cr
& \int_3^\infty {\frac{1}{{x\sqrt {{x^2} - 9} }}} dx = \mathop {\lim }\limits_{a \to {3^ + }} \int_a^4 {\frac{1}{{x\sqrt {{x^2} - 9} }}} dx + \mathop {\lim }\limits_{b \to \infty } \int_4^b {\frac{1}{{x\sqrt {{x^2} - 9} }}} dx \cr
& {\text{Integrate, recall that }}\int {\frac{{dx}}{{x\sqrt {{x^2} - {a^2}} }} = \frac{1}{a}\operatorname{arcsec} \left( {\frac{x}{a}} \right) + C} \cr
& = \mathop {\lim }\limits_{a \to {3^ + }} \left[ {\frac{1}{3}\operatorname{arcsec} \left( {\frac{x}{3}} \right)} \right]_a^4 + \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{3}\operatorname{arcsec} \left( {\frac{x}{3}} \right)} \right]_4^b \cr
& = \frac{1}{3}\mathop {\lim }\limits_{a \to {3^ + }} \left[ {\operatorname{arcsec} \left( {\frac{4}{3}} \right) - \operatorname{arcsec} \left( {\frac{a}{3}} \right)} \right] \cr
& + \frac{1}{3}\mathop {\lim }\limits_{b \to \infty } \left[ {\operatorname{arcsec} \left( {\frac{b}{3}} \right) - \operatorname{arcsec} \left( {\frac{4}{3}} \right)} \right] \cr
& {\text{Evaluate the limits}} \cr
& = \frac{1}{3}\operatorname{arcsec} \left( {\frac{4}{3}} \right) - \frac{1}{3}\operatorname{arcsec} \left( {\frac{{{3^ + }}}{3}} \right) + \frac{1}{3}\operatorname{arcsec} \left( {\frac{\infty }{3}} \right) \cr
& - \frac{1}{3}\operatorname{arcsec} \left( {\frac{4}{3}} \right) \cr
& = - \frac{1}{3}\operatorname{arcsec} \left( {\frac{{{3^ + }}}{3}} \right) + \frac{1}{3}\operatorname{arcsec} \left( {\frac{\infty }{3}} \right) \cr
& = - \frac{1}{3}\left( 0 \right) + \frac{1}{3}\left( {\frac{\pi }{2}} \right) \cr
& = \frac{\pi }{6} \cr} $$
