Answer
$$2{x^2} - 4x + 3\ln \left| {2x - 1} \right| - \frac{1}{{2x - 1}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{16{x^3}}}{{4{x^2} - 4x + 1}}} dx \cr
& {\text{By long division}} \cr
& \frac{{16{x^3}}}{{4{x^2} - 4x + 1}} = 4x + 4 + \frac{{12x - 4}}{{4{x^2} - 4x + 1}} \cr
& \int {\left( {4x - 4} \right)} dx + \int {\frac{{12x - 4}}{{4{x^2} - 4x + 1}}} dx \cr
& \frac{{12x - 4}}{{4{x^2} - 4x + 1}} = \frac{{12x - 4}}{{{{\left( {2x - 1} \right)}^2}}} \cr
& {\text{Decompose the integrand }}\frac{{12x - 4}}{{{{\left( {2x - 1} \right)}^2}}}{\text{ into partial fractions}} \cr
& \frac{{12x - 4}}{{{{\left( {2x - 1} \right)}^2}}} = \frac{A}{{2x - 1}} + \frac{B}{{{{\left( {2x - 1} \right)}^2}}} \cr
& {\text{Multiply by }}{\left( {2x - 1} \right)^2}{\text{ and simplify}} \cr
& 12x - 4 = A\left( {2x - 1} \right) + B \cr
& 12x - 4 = 2Ax - 2A + B \cr
& {\text{Equate coefficients}} \cr
& 12 = 2A,\,\,\,\,\,A = 6,\,\,\,\,\,\, - 2A + B = - 4,\,\,\,\,\,\,B = 2\,\, \cr
& {\text{Replace the coefficients}} \cr
& \int {\left( {4x - 4} \right)} dx + \int {\left( {\frac{6}{{2x - 1}} + \frac{2}{{{{\left( {2x - 1} \right)}^2}}}} \right)} dx \cr
& {\text{Integrate}} \cr
& = 2{x^2} - 4x + 3\ln \left| {2x - 1} \right| - \frac{1}{{2x - 1}} + C \cr} $$
