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