Answer
$$\frac{1}{{3\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{\sqrt 3 x}}{3}} \right) + C $$
Work Step by Step
$$\eqalign{
& \int {\frac{{dx}}{{9 + 3{x^2}}}} \cr
& {\text{write }}9 + 3{x^2}{\text{ as }}{\left( 3 \right)^2} + {\left( {\sqrt 3 x} \right)^2} \cr
& {\text{use the substitution method}}{\text{.}} \cr
& u = \sqrt 3 x,{\text{ so that }}du = \sqrt 3 dx,\,\,\,\,\,\,dx = \frac{{du}}{{\sqrt 3 }} \cr
& {\text{then}} \cr
& \int {\frac{{dx}}{{9 + 3{x^2}}}} = \int {\frac{{du/\sqrt 3 }}{{{{\left( 3 \right)}^2} + {u^2}}}} \cr
& = \frac{1}{{\sqrt 3 }}\int {\frac{{du}}{{{{\left( 3 \right)}^2} + {u^2}}}} \cr
& {\text{intgrate by using the formula }}\int {\frac{{du}}{{{a^2} + {u^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr
& {\text{with }}a = 3 \cr
& = \frac{1}{{\sqrt 3 }}\left( {\frac{1}{3}{{\tan }^{ - 1}}\left( {\frac{u}{3}} \right)} \right) + C \cr
& {\text{write in terms of }}x;{\text{ replace}}\sqrt 3 x{\text{ for }}u \cr
& = \frac{1}{{3\sqrt 3 }}{\tan ^{ - 1}}\left( {\frac{{\sqrt 3 x}}{3}} \right) + C \cr} $$
