Answer
$$x{\tan ^{ - 1}}\left( {3x} \right) - \frac{1}{6}\ln \left( {1 + 9{x^2}} \right) + C$$
Work Step by Step
$$\eqalign{
& \int {{{\tan }^{ - 1}}\left( {3x} \right)} dx \cr
& {\text{substitute }}u = {\tan ^{ - 1}}\left( {3x} \right),{\text{ }}du = \frac{3}{{1 + {{\left( {3x} \right)}^2}}}dx \cr
& du = \frac{3}{{1 + 9{x^2}}}dx \cr
& dv = dx, \cr
& v = x \cr
& {\text{using integration by parts}} \cr
& \int {udv} = uv - \int {vdu} \cr
& {\text{, we have}} \cr
& \int {{{\tan }^{ - 1}}\left( {3x} \right)} dx = x{\tan ^{ - 1}}\left( {3x} \right) - \int {\left( x \right)\left( {\frac{3}{{1 + 9{x^2}}}} \right)dx} \cr
& \int {{{\tan }^{ - 1}}\left( {3x} \right)} dx = x{\tan ^{ - 1}}\left( {3x} \right) - \int {\frac{{3x}}{{1 + 9{x^2}}}dx} \cr
& {\text{find antiderivative}}{\text{, }}t = 1 + 9{x^2}{\text{ }}dt = 18xdx \cr
& \int {{{\tan }^{ - 1}}\left( {3x} \right)} dx = x{\tan ^{ - 1}}\left( {3x} \right) - \frac{3}{{18}}\int {\frac{{dt}}{t}} \cr
& \int {{{\tan }^{ - 1}}\left( {3x} \right)} dx = x{\tan ^{ - 1}}\left( {3x} \right) - \frac{1}{6}\ln \left| t \right| + C \cr
& {\text{replace }}t = 1 + 9{x^2} \cr
& \int {{{\tan }^{ - 1}}\left( {3x} \right)} dx = x{\tan ^{ - 1}}\left( {3x} \right) - \frac{1}{6}\ln \left| {1 + 9{x^2}} \right| + C \cr
& \int {{{\tan }^{ - 1}}\left( {3x} \right)} dx = x{\tan ^{ - 1}}\left( {3x} \right) - \frac{1}{6}\ln \left( {1 + 9{x^2}} \right) + C \cr} $$
