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Recursive integral using integration by pars

  1. Jan 31, 2012 #1
    First excuse my bad english on math subjects. I'm working on it.

    How can I integrate by parts:
    [tex] I_{m}=\int\frac{1}{(x^2+a^2)^m}\,dx[/tex]

    I need to find a recursive form,
    But I can't find the right g' and f to get this done...

    I've tried
    [tex] g'=1 \quad\,\quad\ f=\frac{1}{(x^2+a^2)^m}[/tex]
    As well as [tex]g' = \frac{1}{(x^2+a^2)}\quad\ →g=arctan(x/a)\ , \quad\ f=\frac{1}{(x^2+a^2)^{m-1}} [/tex]

    But on the next integral of g*f ' , I can't find any way to simplify it to Im-1 or another integration by parts that will lead somewhere.

    Which g' and f should I pick for this integral then? Thanks in advance..
     
    Last edited: Jan 31, 2012
  2. jcsd
  3. Jan 31, 2012 #2

    tiny-tim

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    welcome to pf!

    hi oferon! welcome to pf! :smile:

    (btw, your english is fine … on this post at least :wink:)

    (i haven't tried it myself :redface:, but …)

    my guess is that the trick is to write it (x2 + a2)/(x2 + a2)m+1,

    and then use x as f or g
     
  4. Jan 31, 2012 #3
    Hi tim, thanks for your kind reply :redface:

    Which "x" do you reffer to saying "and then use x as f or g" ?

    I've tried the trick you suggested using:

    [tex] g'=x^2+a^2 \quad\, \quad\ f=\frac{1}{(x^2+a^2)^{m+1}} [/tex]

    but all I get is:

    [tex] \frac{arctan(\frac{x}{a})}{(x^2+a^2)^{m+1}} - ∫arctan(\frac{x}{a})*\frac{-2x(m+1)}{(x^2+a^2)^{m+2}}[/tex]

    And now I'm stuck all over again..
     
  5. Jan 31, 2012 #4

    tiny-tim

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    hi oferon! :smile:

    hint: x2/(x2 + a2) = x(x/(x2 + a2)) :wink:
     
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