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Thread: quick integration question...

  1. #1
    Junior Member
    Joined
    Oct 2007
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    51

    quick integration question...

    how does

    $\displaystyle

    \int \frac{-2x+1}{{x^2}+x+1}dx

    $

    become


    $\displaystyle \int \frac{-2x-1}{{x^2}+x+1}dx\
    + 2 \int \frac{dx}{(x+{(1/2)^{2})}+(3/4)}
    $

    i got the completing the squares part but how does it turn to -1 and where did the 2 infront of the second integral come from...
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    Santiago, Chile
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    $\displaystyle \int {\frac{{1 - 2x}}
    {{x^2 + x + 1}}\,dx} = \int {\frac{{2 - \left( {x^2 + x + 1} \right)'}}
    {{x^2 + x + 1}}\,dx} .$

    Now splitting the original integral into two pieces, the second one is ready. (Since there's a hidden logarithm there, you may remove the absolute value bars for $\displaystyle x^2+x+1,$ 'cause it's always positive.) As for the first piece:

    $\displaystyle \int {\frac{2}
    {{x^2 + x + 1}}\,dx} = \int {\frac{8}
    {{4x^2 + 4x + 4}}\,dx} = \int {\frac{8}
    {{(2x + 1)^2 + 3}}\,dx} ,$

    and this is an arctangent, let $\displaystyle u=2x+1$ and proceed from there.
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