# Did I integrate this correctly?

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.

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It might be useful to you to check out this guide on mathjax to make your questions look nice and pretty :D – DanZimm Jun 17 '14 at 12:33
Why are you asking this? Can't you simply differentiate the supposed result and check whehter you get the functions that was integrated?? – DonAntonio Jun 17 '14 at 12:34
@user151764, you have done write but not taken constant $2/21$ – lavkush Jun 17 '14 at 12:34
There are some online calculators to help finding integrals, limits, series, or derivatives. – João Jun 17 '14 at 12:37
What does that have to do with anything at all, @DanZimm ? This is indefinite integration = anti-differentiation. – DonAntonio Jun 17 '14 at 12:39

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 $\dfrac 23$: $$\dfrac 23\cdot \frac{u^{7}}{7} + c = \dfrac 2{21}(x^3 - 4)^7 + c$$

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Okay thank you I see what I did wrong. I forgot to multiply by 2/3 to make du/dx equal 2x^2. – user151764 Jun 17 '14 at 12:39
You're welcome! – amWhy Jun 17 '14 at 12:41

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}

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Thanks mate really helpful! – user151764 Jun 17 '14 at 12:40
@user151764 You're welcome. $\ddot\smile$ – Tunk-Fey Jun 17 '14 at 12:40