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  #151    
Old September 19th, 2011 (4:12 AM).
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Overlord Drakow Overlord Drakow is offline

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    I do not think so. I gave it some thought but nothing springs to mind. That is one seriously nasty integral though ¬___________¬

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      #152    
    Old September 19th, 2011 (4:58 AM).
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    Renii Renii is offline
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      I know.

      I just hope they aren't gonna give questions like that on the test :/

        #153    
      Old September 20th, 2011 (2:26 AM).
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      Overlord Drakow Overlord Drakow is offline

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        I've been doing some more thinking on your problem and have come up with a possible simplification.

        You could try the substitution t = x^k^-1, where k is any real number.

        What that does is inverse the power so you still end up with t^5 and t^4 on the denominator but it should simplify the numerator and make the integrand easier to tackle. I haven't run through the maths or anything but if you want, try it and see.

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          #154    
        Old September 20th, 2011 (3:27 AM).
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        Renii Renii is offline
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          You mean take t = x^(1/20)? (t = x^20^-1 ?)

          I tried that, gives a long answer. It requires applying the binomial theorem with a power of 15)

          Or something else like t = 1/x^k? How? :/

            #155    
          Old September 21st, 2011 (1:45 PM). Edited September 22nd, 2011 by Overlord Drakow.
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          Overlord Drakow Overlord Drakow is offline

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            I meant what I typed out exactly.

            t = x^k^-1

            So for the first term you have x^(1/5) so the transformation gives t^5 and the second term goes from x^(1/4) to t^4

            If that makes sense.

            Edit: The transformation inverses the power of x or in other words you flip the fractions over.

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