Answer
$$\frac{1}{2}\ln \left| {3x + \sqrt {9{x^2} - 25} } \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{\sqrt {9{x^2} - 25} }}} \cr
& = \int {\frac{{dx}}{{\sqrt {{{\left( {3x} \right)}^2} - {{\left( 5 \right)}^2}} }}} \cr
& {\text{substitute 3}}x = 5\sec \theta ,{\text{ 2}}dx = 5\sec \theta \tan \theta d\theta \cr
& = \int {\frac{{\frac{5}{2}\sec \theta \tan \theta d\theta }}{{\sqrt {{{\left( {5\sec \theta } \right)}^2} - {{\left( 5 \right)}^2}} }}} \cr
& {\text{simplify}} \cr
& = \frac{5}{2}\int {\frac{{\sec \theta \tan \theta d\theta }}{{\sqrt {25{{\sec }^2}\theta - 25} }}} \cr
& = \frac{5}{2}\int {\frac{{\sec \theta \tan \theta d\theta }}{{5\sqrt {{{\sec }^2}\theta - 1} }}} \cr
& = \frac{1}{2}\int {\frac{{\sec \theta \tan \theta d\theta }}{{\tan \theta }}} \cr
& = \frac{1}{2}\int {\sec \theta } d\theta \cr
& {\text{find the antiderivative}} \cr
& = \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right| + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{2}\ln \left| {\frac{{3x}}{5} + \frac{{\sqrt {9{x^2} - 25} }}{5}} \right| + C \cr
& \log \,\,{\text{properties}} \cr
& = \frac{1}{2}\ln \left| {3x + \sqrt {9{x^2} - 25} } \right| + C \cr} $$
