Answer
$$\frac{1}{3}{\tan ^{ - 1}}\frac{{3\sqrt x }}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{\sqrt x \left( {9x + 4} \right)}}dx} ,\,\,\,\,u = 3\sqrt x \cr
& = \int {\frac{1}{{\sqrt x \left( {{{\left( {3\sqrt x } \right)}^2} + 4} \right)}}dx} \cr
& {\text{Using the given substitution}} \cr
& u = 3\sqrt x ,\,\,\,\,\,\,\,du = \frac{3}{{2\sqrt x }}dx,\,\,\,\,\,\,dx = \frac{{2\sqrt x }}{3}du \cr
& {\text{write in terms of }}u \cr
& \int {\frac{1}{{\sqrt x \left( {{{\left( {3\sqrt x } \right)}^2} + 4} \right)}}dx} = \int {\frac{1}{{\sqrt x \left( {{u^2} + 4} \right)}}\left( {\frac{{2\sqrt x }}{3}du} \right)} \cr
& = \int {\frac{1}{{\left( {{u^2} + 4} \right)}}\left( {\frac{2}{3}du} \right)} \cr
& = \frac{2}{3}\int {\frac{1}{{{u^2} + 4}}du} \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral}} \cr
& {\text{By formula 68}} \cr
& \left( {68} \right):\,\,\,\,\,\,\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\frac{u}{a} + C} \cr
& {\text{take }}a = 2 \cr
& \frac{2}{3}\int {\frac{1}{{{u^2} + 4}}du} = \frac{2}{{3\left( 2 \right)}}{\tan ^{ - 1}}\frac{u}{2} + C \cr
& {\text{simplifying}} \cr
& \frac{2}{3}\int {\frac{1}{{{u^2} + 4}}du} = \frac{1}{3}{\tan ^{ - 1}}\frac{u}{2} + C \cr
& {\text{write in terms of }}x{\text{ replace }}\cos 2x{\text{ for }}u \cr
& = \frac{1}{3}{\tan ^{ - 1}}\frac{{3\sqrt x }}{2} + C \cr} $$
