Answer
$$\frac{1}{{12}}\ln \left| {\frac{{3 + 2x}}{{3 - 2x}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{9 - 4{x^2}}}} dx \cr
& = \int {\frac{1}{{9 - {{\left( {2x} \right)}^2}}}} dx \cr
& {\text{Let }}u = 2x,{\text{ }}du = 2dx,{\text{ }}\frac{{du}}{2} = dx \cr
& {\text{Substitute}} \cr
& \int {\frac{1}{{9 - {{\left( {2x} \right)}^2}}}} dx = \int {\frac{1}{{9 - {u^2}}}\left( {\frac{1}{2}} \right)} du \cr
& = \frac{1}{2}\int {\frac{1}{{9 - {u^2}}}} du \cr
& {\text{Using the rules of theorem 5}}{\text{.20 }}\left( {{\text{Page 389}}} \right) \cr
& = \frac{1}{2}\left( {\frac{1}{{2\left( 3 \right)}}\ln \left| {\frac{{3 + u}}{{3 - u}}} \right|} \right) + C \cr
& = \frac{1}{{12}}\ln \left| {\frac{{3 + u}}{{3 - u}}} \right| + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{{12}}\ln \left| {\frac{{3 + 2x}}{{3 - 2x}}} \right| + C \cr} $$
