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Double Integration

  1. Sep 27, 2006 #1
    [tex]\int \frac{x^2 - y^2}{(x^2 + y^2)^2} dx[/tex]

    Above is an integration involved in a double integration, I know the answers via TI-89, but I am trying to find out how to get them :frown: I have tried trig substitution, u sub, integration by parts, etc. etc. And I am out of ideas. Can anyone please help?

  2. jcsd
  3. Sep 27, 2006 #2


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    Homework Helper

    Do you know partial fractions?
  4. Sep 27, 2006 #3
    First of all, label your original integral as, say, [itex]I[/itex]

    [tex]I=\int \frac{x^2 - y^2}{(x^2 + y^2)^2} dx[/tex]

    Then split the integral into two seperate integrals, one where [itex]x^2[/itex] is in the numerator, and other where [itex]y^2[/itex] is in the numerator.

    With the first integral (the numerator [itex]x^2[/itex] one), integrate this by parts with the aim of maxing the denominator of your integral [itex](x^2 + y^2)[/itex].

    After you have done this, look at the complete expression for [itex]I[/itex] (substituting into this the integration-by-parts you just carried out).The integrals which remain (the one involving [itex]y^2[/itex] in the numerator, and the integral which is left from integration by parts), combine them both back into one single integral, with the integrand expressed as a single fraction. Compare this integral with the the expression for [itex]I[/itex] on the first line (i.e. the expression above). You should then have an algebraic equation in [itex]I[/itex], which you can solve.
  5. Sep 27, 2006 #4
    Do a trig substitution with x = y Tan[theta], dx = y (Sec[theta])^2. This will reduce your function to something resembling a trig identity that can easily be integrated.
  6. Sep 27, 2006 #5
    Partial fractioning is probably the easiest way as StatusX suggested.
  7. Sep 27, 2006 #6
    Yea I tried that but you end up with

    [tex]x^2 - y^2 = A(x^2 + Y^2) + B(x^2 + Y^2)[/tex]

    And when you set x^2 = -y^2 you would end up with -2y^2 =0 which really isn't a helpful expression :(
  8. Sep 27, 2006 #7


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    Why would you set x^2=-y^2? You need to write:

    [tex]\frac{x^2-y^2}{(x^2+y^2)^2} = \frac{Ax+B}{x^2+y^2}+\frac{Cx+D}{(x^2+y^2)^2}[/tex]

    Then solve for A,B,C,D.
    Last edited: Sep 27, 2006
  9. Sep 27, 2006 #8
    There's a quicker way to manipulate the integrand into a form which ia easier to integrate if you say that

    (x^2 + y^2) = (x+iy)(x-iy)


    x^2 - y^2 = \frac{1}{2}\left[(x+iy)^2 + (x-iy)^2\right]


    \frac{x^2 - y^2}{(x^2 + y^2)^2} = \frac{1}{2}\frac{(x+iy)^2 + (x-iy)^2}{(x+iy)^2(x-iy)^2}
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