Answer
$$\frac{1}{4}\ln \left| {x - 1} \right| + \frac{3}{4}\ln \left| {x + 1} \right| + \frac{1}{{2\left( {x + 1} \right)}} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}dx}}{{\left( {x - 1} \right)\left( {{x^2} + 2x + 1} \right)}}} \cr
& {\text{Decompose the integrand }}\frac{{{x^2}}}{{\left( {x - 1} \right)\left( {{x^2} + 2x + 1} \right)}}{\text{ into partial fractions}} \cr
& \frac{{{x^2}}}{{\left( {x - 1} \right)\left( {{x^2} + 2x + 1} \right)}} = \frac{{{x^2}}}{{\left( {x - 1} \right){{\left( {x + 1} \right)}^2}}} = \frac{A}{{x - 1}} + \frac{B}{{x + 1}} + \frac{C}{{{{\left( {x + 1} \right)}^2}}} \cr
& {\text{Multiply by }}\left( {x - 1} \right){\left( {x + 1} \right)^2}{\text{ and simplify}} \cr
& {x^2} = A{\left( {x + 1} \right)^2} + B\left( {x + 1} \right)\left( {x - 1} \right) + C\left( {x - 1} \right) \cr
& {x^2} = A{x^2} + 2Ax + A + B{x^2} - B + Cx - C \cr
& {\text{Group terms}} \cr
& {x^2} = \left( {A{x^2} + B{x^2}} \right) + \left( {2Ax + Cx} \right) + A - B - C \cr
& {\text{Equating coefficients, we get}} \cr
& A + B = 1,\,\,\,2A + C = 0,\,\,\,A - B - C = 0 \cr
& {\text{Solving the system of linear equation by a calculator, we obtain}} \cr
& A = \frac{1}{4},\,\,\,B = \frac{3}{4},\,\,C = - \frac{1}{2},\,\, \cr
& {\text{Therefore}}{\text{,}} \cr
& \frac{{{x^2}}}{{\left( {x - 1} \right)\left( {{x^2} + 2x + 1} \right)}} = \frac{{1/4}}{{x - 1}} + \frac{{3/4}}{{x + 1}} + \frac{{ - 1/2}}{{{{\left( {x + 1} \right)}^2}}} \cr
& \int {\frac{{{x^2}dx}}{{\left( {x - 1} \right)\left( {{x^2} + 2x + 1} \right)}}} = \int {\left( {\frac{{1/4}}{{x - 1}} + \frac{{3/4}}{{x + 1}} + \frac{{ - 1/2}}{{{{\left( {x + 1} \right)}^2}}}} \right)} dx \cr
& {\text{Integrating, we get}} \cr
& = \frac{1}{4}\ln \left| {x - 1} \right| + \frac{3}{4}\ln \left| {x + 1} \right| + \frac{1}{{2\left( {x + 1} \right)}} + C \cr} $$
