Answer
$$f\left( x \right) = 4\ln \left| {x - 1} \right| - {x^2} + 7$$
Work Step by Step
$$\eqalign{
& f''\left( x \right) = - \frac{4}{{{{\left( {x - 1} \right)}^2}}} - 2 \cr
& {\text{Integrate}} \cr
& f'\left( x \right) = \int {\left[ { - \frac{4}{{{{\left( {x - 1} \right)}^2}}} - 2} \right]dx} \cr
& f'\left( x \right) = \frac{4}{{x - 1}} - 2x + C{\text{ }}\left( {\bf{1}} \right) \cr
& {\text{Using the initial condition }}f'\left( 2 \right) = 0 \cr
& 0 = \frac{4}{{2 - 1}} - 2\left( 2 \right) + C \cr
& 0 = 4 - 4 + C \cr
& C = 0 \cr
& {\text{Substituting }}C{\text{ into }}\left( {\bf{1}} \right) \cr
& f'\left( x \right) = \frac{4}{{x - 1}} - 2x \cr
& {\text{Integrate}} \cr
& f\left( x \right) = \int {\left( {\frac{4}{{x - 1}} - 2x} \right)} dx \cr
& f\left( x \right) = 4\ln \left| {x - 1} \right| - {x^2} + C{\text{ }}\left( {\bf{2}} \right) \cr
& {\text{Using the initial condition }}f\left( 2 \right) = 3 \cr
& 3 = 4\ln \left| {2 - 1} \right| - {\left( 2 \right)^2} + C \cr
& 3 = - 4 + C \cr
& C = 7 \cr
& {\text{Substituting }}C{\text{ into }}\left( {\bf{2}} \right) \cr
& f\left( x \right) = 4\ln \left| {x - 1} \right| - {x^2} + 7 \cr} $$
