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