Answer
$$ = \frac{1}{{6\sqrt 3 }}\ln \left| {\frac{{\sqrt 3 + 3x}}{{\sqrt 3 - 3x}}} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{1}{{3 - 9{x^2}}}} dx{\text{ can be written as}} \cr
& = \int {\frac{1}{{3 - {{\left( {3x} \right)}^2}}}} dx \cr
& {\text{Let }}u = 3x,{\text{ }}du = 3dx,{\text{ }}dx = \frac{1}{3}du \cr
& {\text{Substituting}} \cr
& \int {\frac{1}{{3 - {{\left( {3x} \right)}^2}}}} dx = \int {\frac{1}{{3 - {u^2}}}\left( {\frac{1}{3}} \right)du} \cr
& = \frac{1}{3}\int {\frac{1}{{{{\left( {\sqrt 3 } \right)}^2} - {u^2}}}du} \cr
& {\text{Use Theorem 5}}{\text{.20 }}\int {\frac{{du}}{{{a^2} - {u^2}}} = \frac{1}{{2a}}\ln \left| {\frac{{a + u}}{{a - u}}} \right| + } C \cr
& = \frac{1}{3}\left( {\frac{1}{{2\left( {\sqrt 3 } \right)}}\ln \left| {\frac{{\sqrt 3 + u}}{{\sqrt 3 - u}}} \right|} \right) + C \cr
& = \frac{1}{{6\sqrt 3 }}\ln \left| {\frac{{\sqrt 3 + u}}{{\sqrt 3 - u}}} \right| + C \cr
& {\text{Write in terms of }}x \cr
& = \frac{1}{{6\sqrt 3 }}\ln \left| {\frac{{\sqrt 3 + 3x}}{{\sqrt 3 - 3x}}} \right| + C \cr} $$
