Answer
$$\frac{\pi }{4}$$
Work Step by Step
$$\eqalign{
& \int_2^\infty {\frac{1}{{x\sqrt {{x^2} - 4} }}dx} \cr
& \frac{1}{{x\sqrt {{x^2} - 4} }}{\text{ has an infinite discontinuity at }}x = 2,{\text{so we can write}} \cr
& \int_2^\infty {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx = \mathop {\lim }\limits_{a \to {2^ + }} \int_a^5 {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx + \mathop {\lim }\limits_{b \to \infty } \int_5^b {\frac{1}{{x\sqrt {{x^2} - 4} }}} dx \cr
& {\text{Integrate, recalling that }}\int {\frac{{dx}}{{x\sqrt {{x^2} - {a^2}} }} = \frac{1}{a}\operatorname{arcsec} \left( {\frac{x}{a}} \right) + C} \cr
& = \mathop {\lim }\limits_{a \to {2^ + }} \left[ {\frac{1}{2}\operatorname{arcsec} \left( {\frac{x}{2}} \right)} \right]_a^5 + \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{2}\operatorname{arcsec} \left( {\frac{x}{2}} \right)} \right]_5^b \cr
& = \frac{1}{2}\mathop {\lim }\limits_{a \to {2^ + }} \left[ {\operatorname{arcsec} \left( {\frac{5}{2}} \right) - \operatorname{arcsec} \left( {\frac{a}{2}} \right)} \right] \cr
& + \frac{1}{2}\mathop {\lim }\limits_{b \to \infty } \left[ {\operatorname{arcsec} \left( {\frac{b}{2}} \right) - \operatorname{arcsec} \left( {\frac{4}{2}} \right)} \right] \cr
& {\text{Evaluate the limits}} \cr
& = \frac{1}{2}\operatorname{arcsec} \left( {\frac{5}{2}} \right) - \frac{1}{2}\operatorname{arcsec} \left( {\frac{{{2^ + }}}{2}} \right) + \frac{1}{2}\operatorname{arcsec} \left( {\frac{\infty }{3}} \right) \cr
& - \frac{1}{2}\operatorname{arcsec} \left( {\frac{5}{2}} \right) \cr
& = - \frac{1}{2}\operatorname{arcsec} \left( {\frac{{{2^ + }}}{2}} \right) + \frac{1}{2}\operatorname{arcsec} \left( {\frac{\infty }{2}} \right) \cr
& = - \frac{1}{2}\left( 0 \right) + \frac{1}{2}\left( {\frac{\pi }{2}} \right) \cr
& = \frac{\pi }{4} \cr} $$
