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sean
guest

Dec0505, 05:44 PM (EST) 

"integration problem"

I am trying to integrate the following function f(x)=(1+x^3)^(1/3) w.r.t. x. I am pretty sure that I have gone astray as my solution does not differentiate back to f(x). Anyway, here are the steps that I took Let u = 1+x^3 then du = 3x^2 dx so we have f(u) = u^(1/3) du/3x^2 To eliminate the x^2, I said that x = (u1)^(1/3) ( I suspect this is my first problem ) and I now have f(u) = u^(1/3) du/3((u1)^(1/3))^2 = 1/3 * u^(1/3) (u1)^(2/3) Let z = (u1)^(2/3) then dz = 2/3 u^(1/3) du ( I suspect this is my second problem ) then f(z)=1/3 u^(1/3) 1/z dz/(2/3 u^(1/3)) = 1/2 1/z This I can integrate ( although I suspect I have hidden something incorrect in integratint 1/z ) to 1/2 ln( z ) Substituting back in I get 1/2 ln ( (u1)^(2/3) ) = 1/2 ln ( ( (1+x^3)  1 ) ^(2/3) ) = 1/2 ln( x^2 ) Any pointers to exactly what I have done wrong or suggestions on how else to approach the integration would be appreciated. 

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alexb
Charter Member
1707 posts 
Dec0505, 06:16 PM (EST) 

1. "RE: integration problem"
In response to message #0

>Let z = (u1)^(2/3) then dz = 2/3 u^(1/3) du ( I suspect >this is my second problem ) Yeap, there is a problem there. Regardless, I do not think you'd be able to find any elementary expression for this integral. There might be the G function involved.


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sean
guest

Dec0605, 11:04 AM (EST) 

2. "RE: integration problem"
In response to message #1

I thought that from Chebyshev, any integral of the form x^u * (1+x^v)^(w) would have a solution in elementary functions iff one of (u+1)/v, w, or w + (u+1)/v was an integer.Oh well, my quest continues. Thanks. 

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sammy
guest

Dec0705, 06:42 AM (EST) 

3. "RE: integration problem"
In response to message #0

Wow. thats an amazing piece of integration you did. I wouldnt have thought of it, how did you stumble on it, not the typical procedure for finding the integrand. But its wrong notwit'standing. the reason is that you cant have two different variables in one equation. Like if you are trying to integrate f(z) you cant stick a u in there temporarily and then neatly cancel them out. basically you can only have f(u) = u du or f(z)= z dz for the integrandand so on. its because of the chain rule when you do u substitution. if you are integrating with respect to z, you cannot stick a u into there and treat it as a variable, and then cancel out. they cancel out only because you find a nice z that will cancel the u's. If you have two variables in the integrand, ie u du and z dz, you would have to start integrating with two integrals, or partial integration, and then you would do integration with respect to either variable. email me. anyways, thats my guess. i am a college student, markov276@yahoo.com (just to check it out manually f(z)= 1/3 u^(1/3) 1/z dz/(2/3 u^(1/3)) but, what is dz. ok let me do this one out, f(z)= 1/3 u^(1/3) 1/z <2/3 u^(1/3) du>/(2/3 u^(1/3)) and f(z) = Ok, so i cancel and i get by plugging in z for us i get f(u)... well working backwards it seems logical. true. )


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JJ
guest

Dec0905, 11:59 AM (EST) 

4. "RE: integration problem"
In response to message #3

The function f(x)=(1+x^3)^(1/3) cannot be integrated in terms of a finite number of elementary functions. A special function, the Gauss hypergeometric function, is involved. Sum {from t=0 to t=x}of((1+t^3)^(1/3))dt = x*F((1/3), (1/3); (4/3); x^3) In MATHEMATICA language : Integrate<1/(1+t^3)^(1/3),{t,0,x}> = x*Hypergeometric2F1<1/3,1/3,4/3,x^3>= 

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