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Homework Help: Variable Change in Basic Integration

  1. Apr 28, 2009 #1
    Could someone please explain to me how to do basic integration using the change variable method?

    I understand how to find "u" in the integrand, and then find du/dx, but I don't really know what's going on after that.

    For example:

    [tex]\int[/tex] [tex]\sqrt{ln(x)}[/tex]/x

    u= ln(x), right?
    and du/dx = 1/x

    I know that the integral is equal to [tex]\sqrt{u}[/tex]= (2/3) ln(x)^(3/2)
    But WHY? What happens to 1/x? Is it ignored becuase it's the derivative of ln(x)?

    Last edited: Apr 28, 2009
  2. jcsd
  3. Apr 28, 2009 #2


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    [tex]\int[/tex] [tex]\sqrt{ln(x)}[/tex]/x
    It should be written as
    [tex]\int[/tex] [tex]\sqrt{ln(x)}[/tex]/x*dx
    lnx = u
    du = 1/x*dx
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