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Help me with my integral skills

  1. Feb 21, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex] \int_0^R \! \frac{2rdr}{(r^2+x^2)^{\frac{3}{2}}} [/tex]
    [tex] \int_0^R \! (r^2+x^2)^{-\frac{3}{2}}}d(r^2) [/tex]

    2. Relevant equations



    3. The attempt at a solution
    how did you get from line 1 to 2
     
  2. jcsd
  3. Feb 21, 2009 #2

    rock.freak667

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    d(r^2)=2rdr
     
  4. Feb 21, 2009 #3
    oh I see... so [tex]d(r^2)[/tex] just basically means taking the derivative of [tex]r^2[/tex] with respect with r?
    I am not familiar with that notation
     
  5. Feb 21, 2009 #4

    Tom Mattson

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    Not quite. [itex]d\left(r^2\right)[/itex] is the differential of [itex]r^2[/itex] with respect to [itex]r[/itex].
     
  6. Feb 21, 2009 #5
    I bet you are! (plus or minus a twist)

    Consider [tex]I = \int (u + a^2)^{-\frac{3}{2}} \, du.[/tex]

    Let [tex]u = r^2,[/tex] so [tex]du = 2r\, dr.[/tex] Then

    [tex]I = \int (r^2 + a^2)^{-\frac{3}{2}} \, 2r \, dr\, \text{!}[/tex]
     
  7. Feb 21, 2009 #6
    thanks for clearing that up
     
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