Answer
$$y=\frac{2}{7}(x-3)^{7/2}+\frac{12}{5}(x-3)^{5/2}+6(x-3)^{3/2}+c$$
Work Step by Step
Since$$ \frac{d y}{d x}=x^{2} \sqrt{x-3}$$
Let $ u^2=x-3\ \ \to\ \ 2udu=dx$, then
\begin{align*}
y&=\int x^{2} \sqrt{x-3}dx\\
&=\int (u^2+3)^2(u)(2u)du\\
&=\int (2u^6+12u^4+18u^2)du\\
&=\frac{2}{7}u^7+\frac{12}{5}u^5+\frac{18}{3}u^3+c\\
&=\frac{2}{7}(x-3)^{7/2}+\frac{12}{5}(x-3)^{5/2}+6(x-3)^{3/2}+c
\end{align*}
