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Indefinite integral

  1. May 6, 2010 #1
    1. The problem statement, all variables and given/known data
    integral of 1/x^2/3(1+x^1/3)

    2. Relevant equations
    integral of 1/x dx = ln|x| + c

    3. The attempt at a solution

    let u= x ^2/3(1+x^1/3)
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. May 6, 2010 #2
    note: the stated relevant equation may actually be 1/u du = ln |u| + C not sure
  4. May 6, 2010 #3


    Staff: Mentor

    This is very ambiguous. What exactly is the integrand?
  5. May 6, 2010 #4
    the problem is 1 divided by x^2/3(1+x^1/3) dx

    the integral is indefinite
  6. May 6, 2010 #5
    i dont think it calls for u substitution on further review
  7. May 6, 2010 #6


    Staff: Mentor

    This is still very ambiguous.
    This is what you wrote, but not what I think you meant.
    [tex]\frac{1}{\frac{x^2 (1 + \frac{x^1}{3})}{3}}[/tex]
  8. May 6, 2010 #7
    the ^ sign stands for "to the power of" but if i express the problem in words it would be "one divided by x to the 2/3 power times (1 plus x to the one thirds power)
  9. May 6, 2010 #8


    Staff: Mentor

    Then you should write the integrand as 1/[x^(2/3)(1+x^(1/3))]. Note the parentheses around the exponents.

    Better yet, here's the LaTeX for your integral:
    [tex]\int \frac{1}{x^{2/3}(1 + x^{1/3})} dx[/tex]

    I would start with an ordinary substitution, u = x1/3. I doubt very much that this will turn into du/u.
  10. May 6, 2010 #9
    Then try u = 1+ x1/3, that should do it. I don't think that's too much help, is it?
  11. May 8, 2010 #10
    if u is 1+ x^1/3 then ln |1+x^1/3| + C is correct?
  12. May 8, 2010 #11


    Staff: Mentor

    Not quite. If u = 1 + x^(1/3), du = dx/(3x^(2/3)). Your answer needs to account for that factor of 3 in the denominator.

    Note that x^1/3 [itex]\neq[/itex] x^(1/3). The first is the same as x/3. When you have fractional exponents USE PARENTHESES!
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