• To: mathgroup at smc.vnet.net
• From: BobHanlon at aol.com
• Date: Sun, 30 Apr 2000 21:13:53 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```expr = Log[(((x - 1)^(1/3) - x^(1/3))^2/((x*(x - 1))^(1/3) + (x - 1)^(2/3) +
x^(2/3)))]/18 -
Sqrt[3]*ArcTan[Sqrt[3]*(2*(x - 1)^(1/3) + x^(1/3)) /(3*x^(1/3))]/9 +
x^(2/3)*(x - 1)^(4/3)/2 + x^(2/3)*(x - 1)^(1/3)/3;

D[expr, x] // FullSimplify // PowerExpand // FullSimplify

(x - 1)^(1/3)*x^(2/3)

Would need to verify conditions for which PowerExpand operation is valid.

In a message dated 4/29/2000 10:36:41 PM, bruck at math.usc.edu writes:

>In article <8e3b5h\$kom at smc.vnet.net>, zeno at magicnet.net wrote:
>
>> I symbolically integrated the function..(x^2*(x-1))^(1/3) with respect
>to
>> x.
>>
>> There is in the answer... Hypergeometric2F1[2/3,2/3,5/3,x]
>>
>> I can do nothing more with that..it just returns it. A Hypergeometric2F1
>> with different parameters like Hypergeometric2F1[2,2,5,x] gives an
>> am using version 3. Is Mathematica unable to compute it?
>>
>> I can get the Integral with out the Hypergeometric function on the
>> TI-92+,
>> (it gives the answer in a different for using Tan, etc.) but I still
>
>> would
>> like to work with the Mathematica answer.
>>
>
>zeno e-mailed me the TI-92+ solution, which translated to Mathematica is
>
>  Log[(((x - 1)^(1/3) - x^(1/3))^2/((x*(x - 1))^(1/3) + (x - 1)^(2/3) +
>
>              x^(2/3)))]/18 -
>  Sqrt[3]*ArcTan[Sqrt[3]*(2*(x - 1)^(1/3) + x^(1/3)) /(3*x^(1/3))]/9 +
>
>  x^(2/3)*(x - 1)^(4/3)/2 + x^(2/3)*(x - 1)^(1/3)/3
>
>(I've removed a couple of Abs from the expression, since Mathematica
>doesn't like to differentiate these; as phrased, it's valid for x >= 1).
>
>I am **impressed**.  While I am unable to coax Mathematica to
>differentiate this with respect to x and simplify the derivative to the
>
>original expression, when I plot the difference on [1,2] all I get is
>the typical roundoff noise.  It seems to be correct.
>

Bob

BobHanlon at aol.com

```

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