Did I integrate this correctly?

Question

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.

Answer 1

Letting $$u = x^3 - 4 \implies du = 3x^2\,dx \iff \frac{du}{3} = x^2\,dx \iff \dfrac 23\,du = 2x^2$$

$$\int 2x^2(x^3 - 4)^6 \,dx = \int (\underbrace{x^3 - 4}_{\large u})^6(\underbrace{2x^2\,dx}_{\large \frac 23 \,du}) = \int u^6 \left(\frac 23 du\right) = \dfrac 23\int u^6\,du$$

So you'll need to multiply your result by $\dfrac 23$: $$\dfrac 23\cdot \frac{u^{7}}{7} + c = \dfrac 2{21}(x^3 - 4)^7 + c$$

Answer 2

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}