1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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)?

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

    rl.bhat

    User Avatar
    Homework Helper

    [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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Variable Change in Basic Integration
Loading...