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Substitution integral

  1. Mar 31, 2004 #1
    [tex]\int\frac{dx}{x^4+1}[/tex]
    What really frustrates me is that I've seen this integral before. I believe it involved some whacky subsitution like [tex]x=e^u[/tex], but no substitution seems to work. Partial fractions just make a mess. Trig subs seem tempting but that 4 screws everything up. Ideas?
     
    Last edited: Mar 31, 2004
  2. jcsd
  3. Mar 31, 2004 #2

    matt grime

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    If you do that whacky substitution, how about then multiplying through to get e^(2u)+e^(-2u) and getting a hyperbolic trig function? Not saying this works.
     
  4. Mar 31, 2004 #3
    [tex]\int\frac{dx}{x^4+1}=\frac{1}{3\sqrt{2}}
    \left(2\arctan(1+\sqrt{2}x)-2\arctan(1-\sqrt{2}x)+\right.[/tex]

    [tex]\left.\log(\sqrt{2}x+x^2+1)-\log(\sqrt{2}x-x^2-1)\right)[/tex]
     
  5. Mar 31, 2004 #4

    Hurkyl

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    You could factor the denominator!
     
  6. Apr 1, 2004 #5
    isn't the integral of 1/x^4+1 just the ln|x^4+1|??? or something like that?
     
  7. Apr 1, 2004 #6

    matt grime

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    No, differentiate log(x^4+1) and you'll see why.
     
  8. Apr 1, 2004 #7
    noooo. i mean, the integral of 1/u is the ln|u|+c and in this case, 1/u = (x^4+1)
     
  9. Apr 1, 2004 #8

    ahrkron

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    In order to do that, you also need to have a full [itex]du[/itex] on your integrand, which would need an [itex]x^3[/itex] term that is not there.
     
  10. Apr 2, 2004 #9

    matt grime

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    Also not that if you're correct that the integral of 1/x^2 is log(x^2) and, obviously -1/x as well, so up to a constant 2logx = -1/x?
     
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