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Arctan integral

  1. Jul 8, 2008 #1

    Does anyone know how The Integrator evaluated this:

    "http://integrals.wolfram.com/index.jsp?expr=1%2F%28%28x%5E2%2Ba%5E2%29*%28x%5E2%2By%5E2%2Ba%5E2%29%5E%281%2F2%29%29&random=false" [Broken]

    Last edited by a moderator: Apr 23, 2017 at 2:05 PM
  2. jcsd
  3. Jul 8, 2008 #2


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    Well..perhaps along these lines:
    [tex]k^{2}\equiv{y}^{2}+a^{2}, x=kSinh(u)\to\sqrt{x^{2}+y^{2}+a^{2}}=k\Cosh(u), dx=kCosh(u)du[/tex], whereby our integral should roughly be something like:
  4. Jul 12, 2008 #3
    bump :wink:
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