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 February 24th, 2013, 06:20 AM #1 Newbie   Joined: Feb 2012 Posts: 14 Thanks: 0 Challenging integral Hello, Here it is: ?sqrt(1+(2/3*x^-1/3)^2)
 February 24th, 2013, 06:33 AM #2 Math Team     Joined: Aug 2012 From: Sana'a , Yemen Posts: 1,177 Thanks: 44 Math Focus: Theory of analytic functions Re: Challanging integral The challenging thing here is trying to read the integral , can you write it in latex ..
 February 24th, 2013, 06:57 AM #3 Member     Joined: Jul 2012 Posts: 60 Thanks: 0 Math Focus: Calculus Re: Challanging integral $\int \sqrt{1+$$\frac{2}{3}x^{-\frac{1}{3}}$$^2}\,dx$ First let's make sure we're interpreting the integral correctly. Does your integral look like the one above?
 February 24th, 2013, 08:35 AM #4 Newbie   Joined: Feb 2012 Posts: 14 Thanks: 0 Re: Challanging integral Yes, that's how it looks.
 February 24th, 2013, 08:41 AM #5 Newbie   Joined: Feb 2012 Posts: 14 Thanks: 0 Re: Challanging integral I also have tracks (don't really know what they are called in english), 4 on top of the integral and 0 below. So after i integrate it cant have x as denominator (divided by 0).
February 24th, 2013, 08:51 AM   #6
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Re: Challanging integral

Quote:
 Originally Posted by Realistonia ?sqrt(1+(2/3*x^-1/3)^2)
No point in creating duplicates : viewtopic.php?f=22&t=38531#p157613

Quote:
 Originally Posted by Realistonia Challanging integral
This is not challenging from any direction. It has a very simple elementary (even polynomial, if I am right) form which can easily be seen from u-substitution. Why don't you try some substitutions (have any experience before in solving these problems?) Come on, show some of your efforts, this is not a forum for just solving homework you realize that, right?

 February 24th, 2013, 09:27 AM #7 Newbie   Joined: Feb 2012 Posts: 14 Thanks: 0 Re: Challanging integral Well i did 9 on my own, don't know how do do that one. I got x as denominator and divided by 0 error. If you say that i should use substitutions then maybe you could give some pointers on what should i substitute ?
 February 24th, 2013, 12:25 PM #8 Math Team     Joined: Jul 2011 From: Texas Posts: 3,031 Thanks: 1620 Re: Challanging integral $\sqrt{1 + \frac{4}{9x^{2/3}}}$ $\sqrt{\frac{9x^{2/3}+4}{9x^{2/3}}}$ $\frac{\sqrt{9x^{2/3}+4}}{3x^{1/3}}$ now, let ... $u= 9x^{2/3}+4$ $du= \frac{6}{x^{1/3}} \, dx$
 February 27th, 2013, 05:42 AM #9 Global Moderator   Joined: Dec 2006 Posts: 21,016 Thanks: 2250 The indefinite integral should be x(1+(2/3*x^(-1/3))^2)^(3/2) + C. If the integration limits are 0 and 4, you can evaluate using a calculator for x = 4.

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