Answer
$$\frac{3}{2}\operatorname{arcsec} \left( {3x} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{3}{{2x\sqrt {9{x^2} - 1} }}dx} \cr
& {\text{Let }}u = 3x,{\text{ }}x = \frac{u}{3},{\text{ }}dx = \frac{1}{3}du \cr
& {\text{Substituting}} \cr
& = \int {\frac{3}{{2\left( {u/3} \right)\sqrt {9{{\left( {u/3} \right)}^2} - 1} }}\left( {\frac{1}{3}} \right)du} \cr
& = \frac{3}{2}\int {\frac{1}{{u\sqrt {{u^2} - 1} }}du} \cr
& {\text{Integrate by tables }}\int {\frac{{du}}{{u\sqrt {{u^2} - 1} }} = \operatorname{arcsec} \left| u \right| + C,{\text{ then}}} \cr
& = \frac{3}{2}\operatorname{arcsec} \left| u \right| + C \cr
& {\text{Write in terms of }}u,{\text{ let }}u = 3x \cr
& = \frac{3}{2}\operatorname{arcsec} \left| {3x} \right| + C \cr
& x > \frac{1}{3},{\text{ then}} \cr
& = \frac{3}{2}\operatorname{arcsec} \left( {3x} \right) + C \cr} $$
