Answer
$$F(x) = \frac{1}{3}x^3-5x^2+25x+C$$
Work Step by Step
Given $$f(x) =(x-5)^2=x^2-10x+25$$
Then by using if $f(x)=x^{\alpha}\ \to\ \ F(x) = \frac{x^{\alpha+1}}{\alpha+1}+C$
\begin{align*}
F(x) &=\frac{1}{3}x^3-\frac{10}{2}x^2+25x+C\\
&= \frac{1}{3}x^3-5x^2+25x+C
\end{align*}
To check
\begin{align*}
F'(x) &=x^2-10x+25\\
&=f(x)
\end{align*}