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Finding the integral using partial fractions

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Solve the integral x/x^2+4x+13


    2. Relevant equations
    I think that you would use partial fractions but I'm not really sure. I know that you need to complete the square on the denominator.


    3. The attempt at a solution
    The completed square would be (x+2)^2+9. I don't know what to do now b/c of the x term on top. Help!
     
  2. jcsd
  3. Feb 22, 2009 #2

    HallsofIvy

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    Since the denominator cannot be factored (in terms of real coefficients), there is no "partial fractions".

    Let u= x+ 2 so x= u-2. Then the fraction becomes
    [tex]\frac{u-2}{u^2+ 9}= \frac{u}{u^2+ 9}-2\frac{1}{u^2+ 9}[/tex]
    Let [itex]v= u^2+ 9[/itex] to integrate the first and the second is an arctangent.
     
  4. Feb 22, 2009 #3
    [itex] \frac{x}{x^2+4x+13} = -\frac{2}{x^2+4x+13} + \frac{x+2}{x^2+4x+13} [/itex]

    [itex] \int \frac{dx}{x^2+a^2} = \frac{1}{a}ArcTan\frac{x}{a} + C[/itex]

    [itex] \int \frac{x'}{x} = Ln|x| + C[/itex]

    That's all you need.

    Edit: Method below alot better.
     
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