Answer
$$2x - \frac{{{x^2}}}{2} + \frac{{2{x^3}}}{3} - \frac{{{x^4}}}{4} + C$$
Work Step by Step
$$\eqalign{
& \int {\left( {1 + {x^2}} \right)\left( {2 - x} \right)} dx \cr
& {\text{multiply}}{\text{, foil method}} \cr
& = \int {\left( {\left( 1 \right)\left( 2 \right) + \left( 1 \right)\left( { - x} \right) + \left( {{x^2}} \right)\left( 2 \right) + \left( {{x^2}} \right)\left( { - x} \right)} \right)} dx \cr
& = \int {\left( {2 - x + 2{x^2} - {x^3}} \right)} dx \cr
& = \int 2 dx - \int x dx + \int {2{x^2}} dx - \int {{x^3}} dx \cr
& {\text{power rule}} \cr
& = 2x - \frac{{{x^{1 + 1}}}}{{1 + 1}} + \frac{{2{x^{2 + 1}}}}{{2 + 1}} - \frac{{{x^{3 + 1}}}}{{3 + 1}} + C \cr
& = 2x - \frac{{{x^2}}}{2} + \frac{{2{x^3}}}{3} - \frac{{{x^4}}}{4} + C \cr
& \cr
& {\text{check by differentiation}} \cr
& = \frac{d}{{dx}}\left[ {2x - \frac{{{x^2}}}{2} + \frac{{2{x^3}}}{3} - \frac{{{x^4}}}{4} + C} \right] \cr
& = 2\left( 1 \right) - \frac{{2x}}{2} + \frac{{2\left( {3{x^2}} \right)}}{3} - \frac{{4{x^3}}}{4} \cr
& = 2 - x + 2{x^2} - {x^3} \cr} $$
