Answer
$$\frac{{\sqrt {9{x^2} - 4} }}{{4x}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{x^2}\sqrt {9{x^2} - 4} }}} \cr
& {\text{substitute 3}}x = 2\sec \theta ,{\text{ }}dx = \frac{2}{3}sec\theta tan\theta d\theta \cr
& = \int {\frac{{\frac{2}{3}sec\theta tan\theta d\theta }}{{{{\left( {\frac{2}{3}\sec \theta } \right)}^2}\sqrt {{{\left( {2\sec \theta } \right)}^2} - 4} }}} \cr
& = \frac{3}{2}\int {\frac{{tan\theta d\theta }}{{\sec \theta \sqrt {4{{\sec }^2}\theta - 4} }}} \cr
& = \frac{3}{4}\int {\frac{{tan\theta d\theta }}{{\sec \theta \sqrt {{{\sec }^2}\theta - 1} }}} \cr
& = \frac{3}{4}\int {\frac{{tan\theta d\theta }}{{\sec \theta \sqrt {{{\tan }^2}\theta } }}} \cr
& = \frac{3}{4}\int {\frac{{d\theta }}{{\sec \theta }}} \cr
& = \frac{3}{4}\int {\cos \theta } d\theta \cr
& = \frac{3}{4}\sin \theta + C \cr
& {\text{write in terms of }}x \cr
& = \frac{3}{4}\left( {\frac{{\sqrt {9{x^2} - 4} }}{{3x}}} \right) + C \cr
& = \frac{{\sqrt {9{x^2} - 4} }}{{4x}} + C \cr} $$
