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Integral with roots on bottom and top

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  1. Dec 30, 2014 #1
    It's the integral of sqrt(x)/(cubed root(x) + 1)
    I tried regular u substitution but that didn't let me get rid of all the x's.
    I also just tried long division but that gave me an answer that didn't match with the actual answer to the problem.

    The actual answer is 6[1/7 x^(7/6) - 1/5 x^(5/6) + 1/3 x^(1/6) - x^(1/6) + arctan(x^(1/6)) ] + C
    How do I go about doing this?
     
  2. jcsd
  3. Dec 30, 2014 #2

    mfb

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    Staff: Mentor

    The roots suggest a substitution of x1/6 as this allows to get rid of both powers of x in a reasonable way.
    I did not check if it works, however. The answer is full of powers of that, so it looks good.
     
  4. Jan 16, 2015 #3

    Svein

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    try setting x=u6. The sqrt(x) = u3, cubed root(x) = u2 and dx/du = 6u5. The rest is left to the student...
     
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