Answer
$$\frac{4}{3}\ln \left| {x - 1} \right| - \frac{2}{{3\left( {x - 1} \right)}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{4x - 2}}{{3{{\left( {x - 1} \right)}^2}}}} dx \cr
& \frac{1}{3}\int {\frac{{4x - 2}}{{{{\left( {x - 1} \right)}^2}}}} dx \cr
& {\text{Decomposing the integrand into partial fractions}} \cr
& \frac{{4x - 2}}{{{{\left( {x - 1} \right)}^2}}} = \frac{A}{{x - 1}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}} \cr
& 4x - 2 = A\left( {x - 1} \right) + B \cr
& 4x - 2 = Ax - A + B \cr
& {\text{Equating coefficients}} \cr
& A = 4 \cr
& - A + B = - 2 \to B = 2 \cr
& \cr
& \frac{{4x - 2}}{{{{\left( {x - 1} \right)}^2}}} = \frac{A}{{x - 1}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}} \cr
& \frac{{4x - 2}}{{{{\left( {x - 1} \right)}^2}}} = \frac{4}{{x - 1}} + \frac{2}{{{{\left( {x - 1} \right)}^2}}} \cr
& \frac{1}{3}\int {\frac{{4x - 2}}{{{{\left( {x - 1} \right)}^2}}}} dx = \frac{1}{3}\int {\left[ {\frac{4}{{x - 1}} + \frac{2}{{{{\left( {x - 1} \right)}^2}}}} \right]} dx \cr
& = \frac{1}{3}\int {\frac{4}{{x - 1}}} dx + \frac{2}{3}\int {\frac{1}{{{{\left( {x - 1} \right)}^2}}}} dx \cr
& {\text{Integrating}} \cr
& = \frac{4}{3}\ln \left| {x - 1} \right| + \frac{2}{3}\left( { - \frac{1}{{x - 1}}} \right) + C \cr
& = \frac{4}{3}\ln \left| {x - 1} \right| - \frac{2}{{3\left( {x - 1} \right)}} + C \cr} $$
