Answer
$$\frac{{{x^3}}}{3} - \frac{3}{2}{x^2} + 2x + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {{x^2} - 3x + 2} \right)} dx \cr
& {\text{distribute the integrand by using the sum rule }} \cr
& \int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]} dx = \int {f\left( x \right)} dx \pm \int {g\left( x \right)} dx \cr
& \cr
& = \int {{x^2}} dx - \int {3xdx} + \int 2 dx \cr
& {\text{use multiple constant rule }}\int {k \cdot f\left( x \right)} dx = k\int {f\left( x \right)} dx \cr
& = \int {{x^2}} dx - 3\int {xdx} + 2\int {dx} \cr
& {\text{use power rule }}\int {{x^x}dx} = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr
& = \frac{{{x^{2 + 1}}}}{{2 + 1}} - 3\left( {\frac{{{x^{1 + 1}}}}{{1 + 1}}} \right) + 2\left( {\frac{{{x^{0 + 1}}}}{{0 + 1}}} \right) + C \cr
& {\text{simplifying}} \cr
& = \frac{{{x^3}}}{3} - 3\left( {\frac{{{x^2}}}{2}} \right) + 2\left( x \right) + C \cr
& = \frac{{{x^3}}}{3} - \frac{3}{2}{x^2} + 2x + C \cr} $$