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Integral of a root polynomial

  1. Jul 10, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\int\frac{dx}{x^{1/2}-x^{1/3}}[/tex]


    2. Relevant equations

    None

    3. The attempt at a solution

    What I did was replace [tex]\int\frac{dx}{x^{1/2}-x^{1/3}}[/tex] with [tex]\int\frac{dx}{u^{3}-u^{2}}[/tex] if u=x[tex]^{1/6}[/tex]. Just to simplify things. And I think that was pointless. Help me out?
     
  2. jcsd
  3. Jul 10, 2009 #2
    Re: Integration

    I would suggest that you continue with u = x1/6. What was the last thing you did?
     
  4. Jul 10, 2009 #3
    Re: Integration

    I used integration by parts and got [tex]\frac{1}{u^{2}(u-1)}[/tex]=[tex]\frac{-1}{u}[/tex]-[tex]\frac{1}{u^{2}}[/tex]+[tex]\frac{1}{u-1}[/tex]. Is this right?
     
  5. Jul 10, 2009 #4

    diazona

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    Homework Helper

    Re: Integration

    Well, first you need to convert the dx to an expression involving du. To do so, start by taking the derivative of [itex]u = x^{1/6}[/itex].
     
  6. Jul 10, 2009 #5
    Re: Integration

    So du=[tex]\frac{x^{-5/6}}{6}[/tex]. Now I'm lost again.
     
  7. Jul 10, 2009 #6
    Re: Integration

    Since you don't have x-5/6 (ignoring the constant 1/6) in the integrand, use u = x1/6 to substitute into du = x-5/6/6 dx (don't forget the dx!) so that you only have the variable u in the integrand.
     
  8. Jul 10, 2009 #7
    Re: Integration

    u=x[tex]^{1/6}[/tex] so u[tex]^{6}[/tex]=x

    du=((u[tex]^{6}[/tex])[tex]^{-5/6}[/tex])/6 dx

    So, du=(u[tex]^{-5}[/tex])/6 dx

    Now what?
     
  9. Jul 10, 2009 #8
    Re: Integration

    You want to get dx by itself, then substitute what it's equal to into your integral. After that you shouldn't have any trouble with it.
     
  10. Jul 10, 2009 #9
    Re: Integration

    Hmm...I entered the integral into my calc and its not even close to what I got. This makes no sense!!!
     
  11. Jul 10, 2009 #10

    diazona

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    Homework Helper

    Re: Integration

    Did you use Bohrok's suggestion? Solve for dx in terms of u and du and substitute it in to the integral, so that the integral is entirely in terms of u and du. In other words, what is [itex]f(u)[/itex] in this expression?
    [tex]\int\frac{\mathrm{d}x}{x^{1/2}-x^{1/3}} = \int f(u) \mathrm{d}u[/tex]
     
  12. Jul 10, 2009 #11
    Re: Integration

    I just got it...

    6[tex]\int[/tex][tex]u^{5}/(u^{2}(u-1)[/tex])du.

    Then I simplified and used long division to get a long complicated answer, but its right. Thank you both!
     
  13. Jul 11, 2009 #12
    Re: Integration

    You did good on the start.
    [tex]
    \int\frac{dx}{u^{3}-u^{2}}
    [/tex] where [tex]u=x^{1/6}[/tex]

    The next step is:

    [tex]\int\frac{dx}{u^2(u-1)}[/tex]

    What would you do next?

    Partial fractions. Then substitution, and finally substitution for x=u^6
     
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