Answer
$$ - \frac{x}{{9\sqrt {4{x^2} - 9} }} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{{{\left( {4{x^2} - 9} \right)}^{3/2}}}}} \cr
& {\text{write as}} \cr
& = \int {\frac{{dx}}{{{{\left( {{{\left( {2x} \right)}^2} - 9} \right)}^{3/2}}}}} \cr
& {\text{substitute }}2x = 3\sec \theta ,{\text{ }}dx = \left( {3/2} \right)\sec \theta \tan \theta d\theta \cr
& = \int {\frac{{\left( {3/2} \right)\sec \theta \tan \theta d\theta }}{{{{\left( {{{\left( {3\sec \theta } \right)}^2} - 9} \right)}^{3/2}}}}} \cr
& = \frac{3}{2}\int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\left( {9{{\sec }^2}\theta - 9} \right)}^{3/2}}}}} \cr
& = \frac{3}{2}\int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\left( {9\left( {{{\sec }^2}\theta - 1} \right)} \right)}^{3/2}}}}} \cr
& = \frac{3}{2}\int {\frac{{\sec \theta \tan \theta d\theta }}{{27{{\left( {{{\sec }^2}\theta - 1} \right)}^{3/2}}}}} \cr
& {\text{identity }}{\sec ^2}\theta - 1 = {\tan ^2}\theta \cr
& = \frac{1}{{18}}\int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\left( {{{\tan }^2}\theta } \right)}^{3/2}}}}} \cr
& = \frac{1}{{18}}\int {\frac{{\sec \theta \tan \theta d\theta }}{{{{\tan }^3}\theta }}} \cr
& = \frac{1}{{18}}\int {\frac{{\sec \theta d\theta }}{{{{\tan }^2}\theta }}} \cr
& = \frac{1}{{18}}\int {\left( {\frac{{{{\cos }^2}\theta }}{{{{\sin }^2}\theta }}} \right)\left( {\frac{1}{{\cos \theta }}} \right)} d\theta \cr
& = \frac{1}{{18}}\int {\left( {\frac{{\cos \theta }}{{{{\sin }^2}\theta }}} \right)} d\theta \cr
& = \frac{1}{{18}}\int {\left( {\frac{{\cos \theta }}{{\sin \theta }}} \right)} \left( {\frac{1}{{\sin \theta }}} \right)d\theta \cr
& = \frac{1}{{18}}\int {\cot \theta } \csc \theta d\theta \cr
& {\text{find antiderivative}} \cr
& = - \frac{1}{{18}}\csc \theta + C \cr
& {\text{write in terms }}of{\text{ }}x{\text{ }}\sec \theta = \frac{{2x}}{3}{\text{ then }}\csc \theta = \frac{{2x}}{{\sqrt {4{x^2} - 9} }} \cr
& = - \frac{1}{{18}}\left( {\frac{{2x}}{{\sqrt {4{x^2} - 9} }}} \right) + C \cr
& = - \frac{x}{{9\sqrt {4{x^2} - 9} }} + C \cr} $$
