Answer
$$\frac{1}{4}{\left( {{{\tan }^{ - 1}}\left( {2x} \right)} \right)^2} + 3\ln \left| {x - 2} \right| - \frac{6}{{x - 2}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{{\left( {x - 2} \right)}^2}{{\tan }^{ - 1}}\left( {2x} \right) - 12{x^3} - 3x}}{{\left( {4{x^2} + 1} \right){{\left( {x - 2} \right)}^2}}}} dx \cr
& {\text{Distribute the numerator}} \cr
& = \int {\left( {\frac{{{{\left( {x - 2} \right)}^2}{{\tan }^{ - 1}}\left( {2x} \right)}}{{\left( {4{x^2} + 1} \right){{\left( {x - 2} \right)}^2}}} - \frac{{12{x^3} + 3x}}{{\left( {4{x^2} + 1} \right){{\left( {x - 2} \right)}^2}}}} \right)} dx \cr
& = \int {\left( {\frac{{{{\left( {x - 2} \right)}^2}{{\tan }^{ - 1}}\left( {2x} \right)}}{{\left( {4{x^2} + 1} \right){{\left( {x - 2} \right)}^2}}} - \frac{{3x\left( {4{x^2} + 1} \right)}}{{\left( {4{x^2} + 1} \right){{\left( {x - 2} \right)}^2}}}} \right)} dx \cr
& = \int {\left( {\frac{{{{\tan }^{ - 1}}\left( {2x} \right)}}{{\left( {4{x^2} + 1} \right)}} - \frac{{3x}}{{{{\left( {x - 2} \right)}^2}}}} \right)} dx \cr
& \cr
& {\text{Integrate }}\int {\frac{{3x}}{{{{\left( {x - 2} \right)}^2}}}dx} \cr
& \frac{{3x}}{{{{\left( {x - 2} \right)}^2}}} = \frac{A}{{x - 2}} + \frac{B}{{{{\left( {x - 2} \right)}^2}}} \cr
& 3x = A\left( {x - 2} \right) + B \cr
& 3x = Ax - 2A + B \cr
& A = 3,\,\,\, - 2A + B = 0,\,\,\,B \cr
& \int {\frac{{{{\tan }^{ - 1}}\left( {2x} \right)}}{{4{x^2} + 1}}dx} = \frac{1}{2}\int t dt \cr
& = \frac{1}{2}\left( {\frac{{{t^2}}}{2}} \right) + C \cr
& = \frac{1}{4}{t^2} + C \cr
& {\text{Replace }}t = {\tan ^{ - 1}}\left( {2x} \right) \cr
& = \frac{1}{4}{\left( {{{\tan }^{ - 1}}\left( {2x} \right)} \right)^2} + C \cr
& \cr
& {\text{Integrate }}\int {\frac{{3x}}{{{{\left( {x - 2} \right)}^2}}}dx} \cr
& \frac{{3x}}{{{{\left( {x - 2} \right)}^2}}} = \frac{A}{{x - 2}} + \frac{B}{{{{\left( {x - 2} \right)}^2}}} \cr
& 3x = A\left( {x - 2} \right) + B \cr
& 3x = Ax - 2A + B \cr
& A = 3,\,\,\, - 2A + B = 0,\,\,\,B = 6 \cr
& \frac{{3x}}{{{{\left( {x - 2} \right)}^2}}} = \frac{3}{{x - 2}} + \frac{6}{{{{\left( {x - 2} \right)}^2}}} \cr
& \int {\frac{{3x}}{{{{\left( {x - 2} \right)}^2}}}dx} = \int {\left( {\frac{3}{{x - 2}} + \frac{6}{{{{\left( {x - 2} \right)}^2}}}} \right)dx} \cr
& {\text{Integrate}} \cr
& = 3\ln \left| {x - 2} \right| - \frac{6}{{x - 2}} + C \cr
& {\text{Then}}{\text{,}} \cr
& \int {\left( {\frac{{{{\tan }^{ - 1}}\left( {2x} \right)}}{{\left( {4{x^2} + 1} \right)}} - \frac{{3x}}{{{{\left( {x - 2} \right)}^2}}}} \right)} dx \cr
& \, = \frac{1}{4}{\left( {{{\tan }^{ - 1}}\left( {2x} \right)} \right)^2} + 3\ln \left| {x - 2} \right| - \frac{6}{{x - 2}} + C \cr} $$
