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 ((xx^3)^4/3)/4/3

April 7th, 2010, 06:26 PM  #4  
Senior Member Joined: Nov 2008 Posts: 265 Thanks: 0  Re: integration Quote:
Edit: Don't know if this helps, but you can also say (x(1x)(1+x))^1/3 But thinking a little more on it, you could also say (x(1x^2))^1/3 let u = 1x^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  

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