Answer
$$\frac{{{x^2}}}{2} + 2x + 3\ln \left| {x - 1} \right| - \frac{1}{{x + 1}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^3}}}{{{x^2} - 2x + 1}}} dx \cr
& {\text{By the long division }}\frac{{{x^3}}}{{{x^2} - 2x + 1}} = x + 2 + \frac{{3x - 2}}{{{x^2} - 2x + 1}} \cr
& \cr
& {\text{Decomposing }}\frac{{3x - 2}}{{{x^2} - 2x + 1}}{\text{ into partial fractions}} \cr
& \frac{{3x - 2}}{{{x^2} - 2x + 1}} = \frac{{3x - 2}}{{{{\left( {x - 1} \right)}^2}}} = \frac{A}{{x - 1}} + \frac{B}{{{{\left( {x - 1} \right)}^2}}} \cr
& \,\,\,{\text{Multiply by }}{\left( {x - 1} \right)^2} \cr
& \,\,\,3x - 2 = A\left( {x - 1} \right) + B \cr
& \,\,\,3x - 2 = Ax - A + B \cr
& \,\,{\text{Equating coefficients }}A = 3,\,\,\, - A + B = - 2,\,\,\,B = 1 \cr
& \,\,\,\frac{{3x - 2}}{{{x^2} - 2x + 1}} = \frac{3}{{x - 1}} + \frac{1}{{{{\left( {x - 1} \right)}^2}}} \cr
& \cr
& x + 2 + \frac{{3x - 2}}{{{x^2} - 2x + 1}} = x + 2 + \frac{3}{{x - 1}} + \frac{1}{{{{\left( {x - 1} \right)}^2}}} \cr
& {\text{Then}}{\text{,}} \cr
& \int {\frac{{{x^3}}}{{{x^2} - 2x + 1}}} dx\int {\left( {x + 2 + \frac{3}{{x - 1}} + \frac{1}{{{{\left( {x - 1} \right)}^2}}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = \frac{{{x^2}}}{2} + 2x + 3\ln \left| {x - 1} \right| - \frac{1}{{x + 1}} + C \cr} $$
