Answer
$$\sqrt {{x^2} - 9} - 3{\sec ^{ - 1}}\left( {\frac{x}{3}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{\sqrt {{x^2} - 9} }}{x}} dx \cr
& {\text{substitute }}x = 3\sec \theta ,{\text{ }}dx = 3\sec \theta \tan \theta d\theta \cr
& = \int {\frac{{\sqrt {\left( {3\sec \theta } \right) - 9} }}{{3\sec \theta }}} \left( {3\sec \theta \tan \theta d\theta } \right) \cr
& = \int {\frac{{\sqrt {9{{\sec }^2}\theta - 9} }}{{3\sec \theta }}} \left( {3\sec \theta \tan \theta d\theta } \right) \cr
& = \int {\frac{{3\sqrt {{{\sec }^2}\theta - 1} }}{{3\sec \theta }}} \left( {3\sec \theta \tan \theta d\theta } \right) \cr
& = \int {\frac{{3\sqrt {{{\tan }^2}\theta } }}{{3\sec \theta }}} \left( {3\sec \theta \tan \theta d\theta } \right) \cr
& = 3\int {{{\tan }^2}\theta } d\theta \cr
& = 3\int {\left( {{{\sec }^2}\theta - 1} \right)} d\theta \cr
& {\text{integrating}} \cr
& = 3\left( {\tan \theta - \theta } \right) + C \cr
& = 3\tan \theta - 3\theta + C \cr
& {\text{write in terms of }}x \cr
& = 3\left( {\frac{{\sqrt {{x^2} - 9} }}{3}} \right) - 3{\sec ^{ - 1}}\left( {\frac{x}{3}} \right) + C \cr
& = \sqrt {{x^2} - 9} - 3{\sec ^{ - 1}}\left( {\frac{x}{3}} \right) + C \cr} $$
