Could someone check if I have the right answer for this? I tried wolfram alpha but it used a long division method instead of u substitution.

$\displaystyle

\begin{array}{lll}

\int x(x-3)^2 \: dx

& = & \int (u+3)u^2 \: du \\

& = & \int u^3 + 3u^2 \: du \\

& = & \frac{u^4}{4} + u^2 + c \\

& = & \frac{(x-3)^4}{4} + (x-3)^2 + c

\end{array}

$