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Integral Trouble

  1. Sep 10, 2005 #1
    [tex] \int \frac {x+4}{x^2+2x+5} [/tex]

    I have no idea where to start on this. I cant see any substitutions that would work. I tried completing the square. I also tried to split up the fraction. It isn't getting any simpler. Any help is appreciated.
  2. jcsd
  3. Sep 10, 2005 #2


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    Integrals of this type can always be split into an "ln-part" and an "arctan-part".

    First, calculate the derivative of the denominator and adjust your nominator to get this and split the rest.

    [tex]\left( {x^2 + 2x + 5} \right)^\prime = 2x + 2[/tex]

    The integral then becomes

    [tex]\int {\frac{{x + 4}}{{x^2 + 2x + 5}}} dx = \frac{1}{2}\int {\frac{{2x + 2 + 6}}{{x^2 + 2x + 5}}} dx = \frac{1}{2}\int {\frac{{2x + 2}}{{x^2 + 2x + 5}}} dx + 3\int {\frac{{dx}}{{x^2 + 2x + 5}}}[/tex]

    Now the first integral simply becomes the ln of the denominator, because that's how we 'made' it, it's now of the form f'(x)/f(x).
    The second one no longer has an x in the nominator and you can complete a square in the denominator to get an arctan.
  4. Sep 10, 2005 #3
    I understand that and it makes sense. I just don't get how you determine that's what you do.
  5. Sep 10, 2005 #4


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    Well, this is a standard way for these integrals.
    When the degree of the nominator isn't lower that the one of the denominator, you can divide first (euclidean division on the polynomials) to get this case again.
  6. Sep 10, 2005 #5
    Alright, I think I got this better now. Thanks.
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