TITLE: Did I integrate this correctly?
QUESTION [5 upvotes]: The question was:
$$\int 2x^2 (x^3-4)^6\ dx$$
My answer was $\dfrac{(x^3-4)^7}{7} + C$.
If my answer is wrong please show me the correct method. The textbook doesn't have answers so I turn to my trusty stackexchange users.

REPLY [5 votes]: Let $u=x^3-4\;\Rightarrow\;du=3x^2\ dx$, then
\begin{align}
\require{cancel}
\int 2x^2(x^3-4)^6\ dx&=\int 2\color{red}{\cancel{\color{black}{x^2}}}u^6\cdot\frac{du}{3\color{red}{\cancel{\color{black}{x^2}}}}\\
&=\frac23\int u^6\ du\\
&=\frac23\cdot\frac17u^7+C\\
&=\frac2{21}(x^3-4)^7+C.
\end{align}