Answer
$$x + 4\ln \left| {x - 2} \right| - \ln \left| {x - 1} \right| + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}}}{{{x^2} - 3x + 2}}} dx \cr
& {\text{The integrand is an improper rational function since the numerator }} \cr
& {\text{has degree 2 and the denominator 2}}{\text{. Thus}}{\text{, using the long division}} \cr
& \,\,\,\,\,\,\frac{{{x^2}}}{{{x^2} - 3x + 2}} = 1 + \frac{{3x - 2}}{{{x^2} - 3x + 2}} \cr
& {\text{The integrand can be expressed as}} \cr
& \int {\left( {1 + \frac{{3x - 2}}{{{x^2} - 3x + 2}}} \right)} dx \cr
& = x + \int {\frac{{3x - 2}}{{{x^2} - 3x + 2}}} dx \cr
& {\text{Decomposing the integrand into partial fractions}} \cr
& {\text{Factor the denominator}} \cr
& \frac{{3x - 2}}{{{x^2} - 3x + 2}} = \frac{{3x - 2}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} \cr
& \frac{{3x - 2}}{{\left( {x - 2} \right)\left( {x - 1} \right)}} = \frac{A}{{x - 2}} + \frac{B}{{x - 1}} \cr
& {\text{Multiplying the equation by }}\left( {x - 2} \right)\left( {x - 1} \right){\text{, we have}} \cr
& 3x - 2 = A\left( {x - 1} \right) + B\left( {x - 2} \right) \cr
& {\text{if we set }}x = 2 \cr
& 3\left( 2 \right) - 2 = A\left( {2 - 1} \right) + B\left( 0 \right) \cr
& A = 4 \cr
& {\text{if we set }}x = 1 \cr
& 3\left( 1 \right) - 2 = A\left( {x - 1} \right) + B\left( {x - 2} \right) \cr
& 1 = A\left( 0 \right) + B\left( {1 - 2} \right) \cr
& B = - 1 \cr
& \cr
& {\text{then}} \cr
& x + \int {\frac{{3x - 2}}{{{x^2} - 3x + 2}}} dx = x + \int {\left( {\frac{4}{{x - 2}} - \frac{1}{{x - 1}}} \right)} dx \cr
& {\text{Integrating}} \cr
& = x + 4\ln \left| {x - 2} \right| - \ln \left| {x - 1} \right| + C \cr} $$
