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Integrating with Partial Fractions - Irreducable

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data
    I need to integrate (2x-5)/(x^2+5x+11)


    2. Relevant equations



    3. The attempt at a solution
    My problem is just finding a formula for an irreducable quadratic. I know if the denominator was x(x^2+1), I would use A/x+(Bx+C)/(x^2+1). I just don't know the formula in this case and I feel like I could solve it if I had a formula.
     
  2. jcsd
  3. Sep 13, 2009 #2
    Rewrite the integrand as

    [tex]\frac{2x - 5 + 10 - 10}{x^2 + 5x + 11} = \frac{2x + 5}{x^2 + 5x + 11} - \frac{10}{x^2 + 5x + 11}.[/tex]

    In the new expression, we can integrate the first by noting that the numerator is the derivative of the denominator. For the second, complete the square and substitute.
     
  4. Sep 13, 2009 #3
    Ok, I sorta understand that, but let's say the numerator was something like 3x-7. Is there something different to do in this case?
     
  5. Sep 13, 2009 #4
    Not really, the basic idea is to ensure that we have some expression where the derivative of the quadratic in the denominator is the numerator.

    If 3x - 7 was in the numerator, then 3x - 7 = (3/2)(2x - 14/3). Hence,

    [tex]\frac{3x - 7}{x^2 + 5x + 11} = \frac{3}{2}\cdot\frac{2x - 14/3 + 29/3 - 29/3}{x^2 + 5x + 11} = \frac{3}{2}\left(\frac{2x + 5}{x^2 + 5x + 11} - \frac{29}{3}\cdot\frac{1}{x^2 + 5x + 11}\right)[/tex]
     
  6. Sep 13, 2009 #5
    That's a good way to do it, remembering that:

    [tex]\int \frac{f'(x)}{f(x)} dx = \log_e (f(x))[/tex]

    and that

    [tex]\int \frac{dx}{x^2+a^2} = \frac{\tan^{-1} (\frac{x}{a}) }{a} [/tex]

    Seeing that the integral can be split in the fashion described is a good start and maybe it's the first step you should look to take when seeing an integrand that is a quotient of two polynomials the numerator being of one less degree than the denominator.
     
  7. Sep 13, 2009 #6
    Thanks! I figured it out!
     
  8. Sep 13, 2009 #7
    I think this would be general for problems like these?

    [tex] \int \frac{Ax+B}{ax^2+bx+c} dx = \frac{A}{2a}\int \frac{2a}{A}.\frac{Ax+B}{ax^2+bx+c} = \frac{A}{2a}\left( \int\frac{2ax+b}{ax^2+bx+c}+\int{\frac{\frac{2aB}{A}-b}{ax^2+bx+c}\right)[/tex]

    [tex] \int \frac{\frac{2aB}{A}-b}{ax^2+bx+c} = \frac{\frac{2aB}{A}-b}{a\sqrt{c-\frac{b^2}{4a^2}}}\tan^{-1}\left(\frac{x+\frac{b}{2a}}{\sqrt{c-\frac{b^2}{4a^2}}}\right)[/tex]

    So

    [tex]\int \frac{Ax+B}{ax^2+bx+c} dx = \frac{A}{2a}\left(log_e(ax^2+bx+c) + \frac{\frac{2aB}{A}-b}{a\sqrt{c-\frac{b^2}{4a^2}}}\tan^{-1}\left(\frac{x+\frac{b}{2a}}{\sqrt{c-\frac{b^2}{4a^2}}}\right)\right)[/tex]
     
  9. Sep 15, 2009 #8
    Where does the 29/3 come from? I see that we want 2x+5 in the numerator. So we would have 3x-7+12-12, right? From there, I get (3/2)(2x-14/3+24/3-24/3). This will give us what we want. I'm just trying to see where 29/3 came from instead of 24/3?
     
  10. Sep 15, 2009 #9
    I think I figured it out. It's because of the 3/2 factored out?
     
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