Answer
$$\frac{1}{6}{\left( {{{\tan }^{ - 1}}\left( {3x} \right)} \right)^2} + \ln \left| {x + 1} \right| + \frac{1}{{x + 1}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\left( {x + 1} \right)}^2}{{\tan }^{ - 1}}\left( {3x} \right) + 9{x^3} + x}}{{\left( {9{x^2} + 1} \right){{\left( {x + 1} \right)}^2}}}} dx \cr
& {\text{Distribute the numerator}} \cr
& = \int {\left( {\frac{{{{\left( {x + 1} \right)}^2}{{\tan }^{ - 1}}\left( {3x} \right)}}{{\left( {9{x^2} + 1} \right){{\left( {x + 1} \right)}^2}}} + \frac{{9{x^3} + x}}{{\left( {9{x^2} + 1} \right){{\left( {x + 1} \right)}^2}}}} \right)} dx \cr
& = \int {\left( {\frac{{{{\left( {x + 1} \right)}^2}{{\tan }^{ - 1}}\left( {3x} \right)}}{{\left( {9{x^2} + 1} \right){{\left( {x + 1} \right)}^2}}} + \frac{{x\left( {9{x^2} + 1} \right)}}{{\left( {9{x^2} + 1} \right){{\left( {x + 1} \right)}^2}}}} \right)} dx \cr
& = \int {\left( {\frac{{{{\tan }^{ - 1}}\left( {3x} \right)}}{{9{x^2} + 1}} + \frac{x}{{{{\left( {x + 1} \right)}^2}}}} \right)} dx \cr
& \cr
& {\text{Integrate }}\int {\frac{{{{\tan }^{ - 1}}\left( {3x} \right)}}{{9{x^2} + 1}}dx} \cr
& {\text{let }}t = {\tan ^{ - 1}}\left( {3x} \right),\,\,\,\,dt = \frac{3}{{9{x^2} + 1}}dx \cr
& \int {\frac{{{{\tan }^{ - 1}}\left( {3x} \right)}}{{9{x^2} + 1}}dx} = \frac{1}{3}\int t dt \cr
& = \frac{1}{3}\left( {\frac{{{t^2}}}{2}} \right) + C \cr
& = \frac{1}{6}{t^2} + C \cr
& {\text{replace }}t = {\tan ^{ - 1}}\left( {3x} \right) \cr
& = \frac{1}{6}{\left( {{{\tan }^{ - 1}}\left( {3x} \right)} \right)^2} + C \cr
& \cr
& {\text{Integrate }}\int {\frac{x}{{{{\left( {x + 1} \right)}^2}}}dx} \cr
& \frac{x}{{{{\left( {x + 1} \right)}^2}}} = \frac{A}{{x + 1}} + \frac{B}{{{{\left( {x + 1} \right)}^2}}} \cr
& x = A\left( {x + 1} \right) + B \cr
& x = Ax + A + B \cr
& A = 1,\,\,\,A + B = 0,\,\,\,\,B = - 1 \cr
& \frac{x}{{{{\left( {x + 1} \right)}^2}}} = \frac{1}{{x + 1}} - \frac{1}{{{{\left( {x + 1} \right)}^2}}} \cr
& \int {\frac{x}{{{{\left( {x + 1} \right)}^2}}}dx} = \int {\left( {\frac{1}{{x + 1}} - \frac{1}{{{{\left( {x + 1} \right)}^2}}}} \right)dx} \cr
& {\text{Integrate}} \cr
& = \ln \left| {x + 1} \right| + \frac{1}{{x + 1}} + C \cr
& {\text{Then}}{\text{,}} \cr
& \, = \frac{1}{6}{\left( {{{\tan }^{ - 1}}\left( {3x} \right)} \right)^2} + \ln \left| {x + 1} \right| + \frac{1}{{x + 1}} + C \cr} $$
