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An integral calculation

  1. Mar 12, 2005 #1
    Hi,
    I have a problem. I can't understand how it becomes [-1/the square root of (z^2+r^2)] from integral (r.dr/(z^2+r^2)^3/2)
    It seems complicated but I couldn't write it different.
    Thanks, Sol
    P.S. The integral goes to R from 0.
     
  2. jcsd
  3. Mar 12, 2005 #2

    Tom Mattson

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    It's a simple u-substitution. Try to do it, and come back if you get stuck.
     
  4. Mar 12, 2005 #3

    arildno

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    [tex]\frac{d}{dr}(-\frac{1}{(z^{2}+r^{2})^{\frac{1}{2}}})=\frac{1}{2}*\frac{1}{(z^{2}+r^{2})^{\frac{3}{2}}}*2r=\frac{r}{(z^{2}+r^{2})^{\frac{3}{2}}}[/tex]
    That's all there is to it, really..
     
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