Answer
$$F(x)=\frac{1}{2}x^4-\frac{2}{9}x^3+\frac{5}{2}x^2+C$$
Work Step by Step
Given $$f(x) =2x^3-\frac{2}{3}x^2+5x$$
Then by using if $f(x)=x^{\alpha}\ \to\ \ F(x) = \frac{x^{\alpha+1}}{\alpha+1}+C$
\begin{align*}
F(x) &=\frac{2x^4}{4}-\frac{2}{3}\frac{x^3}{3}+\frac{5}{2}x^2+C\\
&=\frac{1}{2}x^4-\frac{2}{9}x^3+\frac{5}{2}x^2+C
\end{align*}
To check
\begin{align*}
F'(x) &=2x^3-\frac{2}{3}x^2+5x\\
&=f(x)
\end{align*}
