Answer
$$\frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{2x}}{3}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{9 + 4{x^2}}}} dx \cr
& {\text{substitute }}u = 2x,{\text{ }}du = 2dx,{\text{ }}a = 3 \cr
& = \int {\frac{1}{{{{\left( 3 \right)}^2} + {{\left( {2x} \right)}^2}}}} dx \cr
& = \int {\frac{{\left( {1/2} \right)}}{{{a^2} + {u^2}}}} dx \cr
& = \frac{1}{2}\int {\frac{1}{{{a^2} + {u^2}}}} dx \cr
& {\text{ use the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}}} = \frac{1}{a}ta{n^{ - 1}}\left( {\frac{u}{a}} \right) + C,{\text{ }}\left( {{\text{see page 468}}} \right) \cr
& = \frac{1}{{2a}}{\tan ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& {\text{write in terms of }}x \cr
& = \frac{1}{{2\left( 3 \right)}}{\tan ^{ - 1}}\left( {\frac{{2x}}{3}} \right) + C \cr
& = \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{2x}}{3}} \right) + C \cr} $$
