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Integration formula

  1. Jan 22, 2008 #1
    Hello all,

    My textbook states the formula [itex]\int\frac{du}{(u^2+\alpha^2)^m}=\frac{1}{2\alpha^2(m-1)}\frac{u}{(u^2+\alpha^2)^{m-1}}+\frac{2m-3}{2\alpha^2(m-1)}\int\frac{du}{(u^2+\alpha^2)^{m-1}}[/itex] but does not provide a proof of this formula. Anyone want to show me how it's derived? I tried integration by parts, which the book gives as the method for deriving the formula, but I couldn't figure it out :-(. Any help?
     
  2. jcsd
  3. Jan 23, 2008 #2

    EnumaElish

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    You may want to try the homework section.
     
  4. Jan 23, 2008 #3
    Why?

    Also any mods can move this if necessary but it isn't homework for a class...
     
  5. Jan 23, 2008 #4

    EnumaElish

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    A "homework-type" question might get a faster response in the HW section.
     
  6. Jan 24, 2008 #5
    You have a point. I don't know how to move it or if that's even possible. If a mod sees this please move it.
     
  7. Jan 24, 2008 #6

    Gib Z

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    An obvious trig substitution converts this integral into a power of cosine integral, which has well known recursion formulae, or otherwise, easier to derive recursion formulae than the original integral.

    EDIT: The hyperbolic substitution converts it into a power of hyperbolic cosine integral, which is even easier to deal with.
     
    Last edited: Jan 24, 2008
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