Đáp án: $B$
Giải thích các bước giải:
Ta có :
$I=\int\dfrac{(x-2)(x^2-x+2)}{x+2}dx$
$\to I=\int\dfrac{x^3-3x^2+4x-4}{x+2}dx$
$\to I=\int\dfrac{(x^3+2x^2)-(5x^2+10x)+(14x+28)-32}{x+2}dx$
$\to I=\int\dfrac{x^2(x+2)-5x(x+2)+14(x+2)-32}{x+2}dx$
$\to I=\int x^2-5x+14-\dfrac{32}{x+2}dx$
$\to I=\dfrac{x^3}{3}-\dfrac{5x^2}{2}+14x-32\ln|x+2|+C$
$\to \int^2_1\dfrac{(x-2)(x^2-x+2)}{x+2}dx=\dfrac{53}{6}-64\ln2+32\ln3$
$\to a=\dfrac{53}{6}, b=-64,c=32\to c>0\to B$
