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Difficult Integral of a Rational Function

  1. Jun 27, 2009 #1
    Hey guys I'm wondering if someone could hep me solve this integral. I've been working at it for a few days now (as part of a project I'm doing over the summer) and have gotten stuck. I think I need to make some substitution but I cant see what it is to make.

    [tex]-\int\frac{dI}{I(R+BI+CI^2)}[/tex]

    I decomposed using partial fractions and reduced it to this:

    [tex]-\frac{1}{R}\int{\frac{dI}{I}+\frac{(CI+B)dI}{R+BI+CI^2}}[/tex]

    I think I need to make another substitution here for the right-hand part of the integral. Simple U substitution doesn't work but I'm not sure of another method that would help. I tried completing the square for the polynomial on the bottom but that didn't seem to help.

    Any help would be really appreciated! Thanks
     
  2. jcsd
  3. Jun 27, 2009 #2
    [tex]
    -\int\frac{d x}{x(R+B x+C x^2)} = -\frac{\frac{2 B \tan ^{-1}\left(\frac{B+2 C x}{\sqrt{4 C
    R-B^2}}\right)}{\sqrt{4 C R-B^2}}+\log (x (B+C x)+R)-2 \log (x)}{2
    R}[/tex]
     
  4. Jun 27, 2009 #3
    Yes I got that answer with mathematica but I need to be able to solve it by hand. Is it possible?
     
  5. Jun 27, 2009 #4

    George Jones

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    [tex]
    \frac{CI + B}{R + BI + CI^2} = \frac{CI + B/2}{R + BI + CI^2} + \frac{B/2}{R + BI + CI^2} = \frac{1}{2} \frac{2CI + B}{R + BI + CI^2} + \frac{B/2}{R + BI + CI^2}
    [/tex]

    Integrate the first term above by inspection or by a simple substitution to get a ln. Complete the square for the denominator of the second term to get an arctan.
     
  6. Jun 28, 2009 #5
    Ok I was able to do it now thanks!
     
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