Answer
$$F(x)=\frac{1}{3}x^3-\frac{3x}{2}^2+2x+C$$
Work Step by Step
Given $$f(x) =x^2-3x+2 $$
Then by using if $f(x)=x^{\alpha}\ \to\ \ F(x) = \frac{x^{\alpha+1}}{\alpha+1}+C$
\begin{align*}
F(x) &=\frac{x^3}{3}-\frac{3x^{2}}{2}+2x+C\\
&=\frac{1}{3}x^3-\frac{3x}{2}^2+2x+C
\end{align*}
To check
\begin{align*}
F'(x) &=x^2-3x+2\\
&=f(x)
\end{align*}
