Answer
$$\dfrac{\pi}{6}$$
Work Step by Step
Recall the formula: $\int \dfrac{du}{u \sqrt {u^2-a^2}}=\dfrac{1}{a} \sec^{-1} \dfrac{x}{a}+c$
Consider $I=\int_3^\infty \dfrac{dx}{x \sqrt {x^2-3^2}}$
or, $=\lim\limits_{a \to 3^{+}} \int_{a}^\infty \dfrac{dx}{x \sqrt {x^2-3^2}} $
or, $=\lim\limits_{a \to 3^{+}} \dfrac{1}{3} \sec^{-1} (\dfrac{x}{3})]_3^\infty$
or, $=\dfrac{1}{3} [\sec^{-1} \dfrac{a}{3} -\sec^{-1} (1) ]$
So, $I=\dfrac{1}{3} (\dfrac{\pi}{2}-0) =\dfrac{\pi}{6}$
