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Math Help - quick integration question...

  1. #1
    Junior Member
    Joined
    Oct 2007
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    51

    quick integration question...

    how does

    <br /> <br />
\int \frac{-2x+1}{{x^2}+x+1}dx<br /> <br />

    become


     \int \frac{-2x-1}{{x^2}+x+1}dx\<br />
+ 2 \int \frac{dx}{(x+{(1/2)^{2})}+(3/4)}<br />

    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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    \int {\frac{{1 - 2x}}<br />
{{x^2  + x + 1}}\,dx}  = \int {\frac{{2 - \left( {x^2  + x + 1} \right)'}}<br />
{{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 x^2+x+1, 'cause it's always positive.) As for the first piece:

    \int {\frac{2}<br />
{{x^2  + x + 1}}\,dx}  = \int {\frac{8}<br />
{{4x^2  + 4x + 4}}\,dx}  = \int {\frac{8}<br />
{{(2x + 1)^2  + 3}}\,dx} ,

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