Hi:

I think I have gone horribly wrong here:

Could someone take a peek please?

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- Mar 1st 2010, 02:11 PM #1

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- Mar 1st 2010, 02:24 PM #2

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Let $\displaystyle u = 5x^4 + 5$.

Then:

$\displaystyle \int 3x^3(5x^4 + 5)^3 \cdot dx =$

$\displaystyle \int 3 \cdot \frac{20}{20} x^3(5x^4 + 5)^3 \cdot dx =$

$\displaystyle \int \frac{3}{20} \cdot 20x^3(5x^4 + 5)^3 \cdot dx =$

$\displaystyle \int \frac{3}{20} \cdot u^3 \cdot du$

- Mar 1st 2010, 02:26 PM #3

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- Mar 1st 2010, 02:28 PM #4

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- Mar 3rd 2010, 07:25 AM #5

- Mar 3rd 2010, 02:59 PM #6

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sorry icemanfan but i couldn't see what your trying to say with that.

anyway i think i made a mistake above by leaving out the root^3

i have adjusted if someone could check please

<img>http://www.mathhelpforum.com/math-help/attachments/calculus/15753d1267660768-integratetion-3-q3.2.gif</img>

- Mar 3rd 2010, 03:03 PM #7

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- Mar 3rd 2010, 05:30 PM #8

- Mar 3rd 2010, 10:19 PM #9

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Hi:

So have I got the final answer here?

If so, I'm a little confused. Is this integration by parts?

Because the way that this is calculated doesn't go with the formula I've been taught and method.

If this is the correct final answer, can somebody explain please?

- Mar 3rd 2010, 10:22 PM #10

- Mar 3rd 2010, 10:26 PM #11

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Great!

Thanks Vita X.

Thanks to everyone who contributed.

Any tips on why I got my method mixed up here?

I think the part I'm confused with is du=20x^3dx and 1/20du=x^3

It seems I am getting confused with integral products.

- Mar 3rd 2010, 10:43 PM #12

- Mar 3rd 2010, 10:51 PM #13

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- Mar 3rd 2010, 10:57 PM #14

- Mar 3rd 2010, 10:59 PM #15