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Thread: Integral w.r.t y alone

  1. #1
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    Integral w.r.t y alone

    Im working on a problem an this comes up $\displaystyle \frac{d\psi}{dy}=\frac{2a(y^2-x^2+a^2)}{((x-a)^2 +y^2)((x-a)^2 +y^2)} $ ho do i integrate something like that to get $\displaystyle atan(\frac{y}{x+a}) - atan(\frac{y}{x-a}) $ ? It seems at the very least unfriendly to partial fractions and just looks tedious
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  2. #2
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    Quote Originally Posted by phycdude View Post
    Im working on a problem an this comes up $\displaystyle \frac{d\psi}{dy}=\frac{2a(y^2-x^2+a^2)}{((x-a)^2 +y^2)((x-a)^2 +y^2)} $ ho do i integrate something like that to get $\displaystyle atan(\frac{y}{x+a}) - atan(\frac{y}{x-a}) $ ? It seems at the very least unfriendly to partial fractions and just looks tedious
    Actually it splits very nicely

    $\displaystyle
    \frac{d\psi}{dy}= \frac{2a(y^2-x^2+a^2)}{((x-a)^2 +y^2)((x-a)^2 +y^2)} = \;\; \frac{x+a}{(x+a)^2+y^2} - \frac{x-a}{(x-a)^2+y^2}
    $.
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  3. #3
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    Quote Originally Posted by Danny View Post
    Actually it splits very nicely

    $\displaystyle
    \frac{d\psi}{dy}= \frac{2a(y^2-x^2+a^2)}{((x-a)^2 +y^2)((x-a)^2 +y^2)} = \;\; \frac{x+a}{(x+a)^2+y^2} - \frac{x-a}{(x-a)^2+y^2}
    $.


    how did u do that?
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  4. #4
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    Quote Originally Posted by Danny View Post
    Actually it splits very nicely

    $\displaystyle
    \frac{d\psi}{dy}= \frac{2a(y^2-x^2+a^2)}{((x-a)^2 +y^2)((x-a)^2 +y^2)} = \;\; \frac{x+a}{(x+a)^2+y^2} - \frac{x-a}{(x-a)^2+y^2}
    $.
    well there seems to be a typo in the denominator, my bad heres the correction
    $\displaystyle
    \frac{d\psi}{dy}= \frac{2a(y^2-x^2+a^2)}{((x+a)^2 +y^2)((x-a)^2 +y^2)} $
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