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Homework Help: General integration formulae

  1. May 21, 2010 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    I was intent on finding a general formula for every case where the constants are real in this integral:

    [tex]\int\frac{dx}{ax^2+bx+c}[/tex]

    But thought I would make things progressive by tackling a seemingly easier problem, mainly:

    [tex]\int\frac{ax+b}{cx^2+dx+e}dx[/tex]


    3. The attempt at a solution

    I let [tex]cx^2+dx+e=u[/tex]

    [tex]du=(2cx+d)dx[/tex]

    Now I needed the numerator to be equivalent to du, so after some manipulation I get

    [tex]ax+b=\frac{a}{2c}(2cx+d+\frac{2bc}{a}-d)[/tex]

    Which made me realize that this problem isn't any easier than the first, since I'm going to have to solve the first anyway...

    [tex]\int\frac{ax+b}{cx^2+dx+e}dx[/tex]

    [tex]=\frac{a}{2c}\left(\int\frac{2cx+d}{u}dx+\int\frac{\frac{2bc}{a}-d}{u}dx\right)[/tex]

    [tex]=\frac{a}{2c}\left(\int\frac{du}{u}+\frac{2bc}{a}\int\frac{dx}{cx^2+dx+e}-d\int\frac{dx}{cx^2+dx+e}\right)[/tex]

    So I guess my question is how do I deal with this integral in the case that the quadratic has no real factors. i.e. b2-4ac<0.
     
    Last edited: May 21, 2010
  2. jcsd
  3. May 21, 2010 #2

    rock.freak667

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    Try completing the square and then use a trig substitution


    EDIT: I mean you can do this in the initial problem as well.
     
  4. May 21, 2010 #3

    Mentallic

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    Oh right, thank you :smile:

    For [tex]\int\frac{dx}{ax^2+bx+c}[/tex] completing the square gives

    [tex]\frac{1}{a}\int\frac{dx}{\left(x+\frac{b}{2a}\right)^2+\frac{4ac-b^2}{4a^2}}[/tex]

    Then letting [tex]x=\frac{-b}{2a}+\frac{\sqrt{4ac-b^2}}{2a}tanu[/tex]

    The answer becomes

    [tex]\frac{2}{\sqrt{4ac-b^2}}.tan^{-1}\left(\frac{2ax+b}{\sqrt{4ac-b^2}}\right)[/tex]

    I was really unsure of myself, and trying the derivative, it was quite amazing to watch everything cancel out to once again become the reciprocal of the general quadratic.

    I'll try it for the other 2 cases as well now.
     
  5. May 21, 2010 #4

    Gib Z

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    That is only valid for when b^2 - 4ac > 0. You need to account for the other case as well.
     
  6. May 21, 2010 #5

    Mentallic

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    Yep, that one is for the case that [tex]b^2-4ac<0[/tex]

    For [tex]b^2-4ac=0[/tex] the result is [tex]\frac{-2}{2ax+b}+c[/tex]

    and for [tex]b^2-4ac>0[/tex], [tex]\frac{1}{\beta -\gamma}ln\left(\frac{x-\beta}{x-\gamma}\right)[/tex], where [itex]\beta,\gamma[/itex] are the two roots of the quadratic.
     
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