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 April 6th, 2010, 06:54 PM #1 Senior Member   Joined: Jul 2009 Posts: 138 Thanks: 0 integration
 April 6th, 2010, 07:03 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,831 Thanks: 2160 I think this integral can't be done in terms of the usual elementary functions.
 April 7th, 2010, 02:36 PM #3 Newbie   Joined: Mar 2010 Posts: 3 Thanks: 0 Re: integration sorry I edited a couple times would it be ((x-x^3)^4/3)/4/3
April 7th, 2010, 06:26 PM   #4
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Re: integration

Quote:
 Originally Posted by nycmets718 sorry I edited a couple times would it be ((x-x^3)^4/3)/4/3
No it wouldn't. The problem is the x - x^3. I can't think of any way to integrate that for now, unless you want to use integration by parts somehow, but that doesn't seem right.

Edit: Don't know if this helps, but you can also say (x(1-x)(1+x))^1/3

But thinking a little more on it, you could also say (x(1-x^2))^1/3

let u = 1-x^2
du = -2x dx

so...-2[(u)^1/3 du](of course don't forget the integral)
then I think you can go from there

 April 8th, 2010, 02:54 PM #5 Senior Member   Joined: Jul 2009 Posts: 138 Thanks: 0 Re: integration
 April 8th, 2010, 06:57 PM #6 Senior Member   Joined: Nov 2008 Posts: 265 Thanks: 0 Re: integration Yeah, I knew I was wrong from the beginning. That's a lot of work for a small problem.
 April 10th, 2010, 07:59 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,831 Thanks: 2160 Well done, rose. Your image is a bit difficult to read, but looks okay, except that the final answer isn't defined for x = 0. However, some simple manipulation can be used to convert it to a form that is defined for x = 0 (and equivalent elsewhere).
 April 11th, 2010, 08:59 AM #8 Newbie   Joined: Apr 2010 Posts: 1 Thanks: 0 Re: integration good thinking
 October 25th, 2017, 01:08 AM #9 Newbie   Joined: Oct 2017 From: Texas Posts: 9 Thanks: 0 That is basic integration. Hint- Check the integration formulas here.

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