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Math Help - another integration

  1. #1
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    another integration

    integrate (x-3)/(x^2 + 2x+4).
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by twilightstr View Post
    integrate (x-3)/(x^2 + 2x+4).
    note that \int \frac {x - 3}{x^2 + 2x + 4}~dx = \int \frac {x + 1}{x^2 + 2x + 4}~dx - \int \frac 4{x^2 + 2x + 3}

    for the first integral, a substitution of u = x^2 + 2x + 4 works.

    for the second integral, note that \frac 4{x^2 + 2x + 4} = \frac 4{(x + 1)^2 + 3} = \frac 43 \cdot \frac 1{\left( \frac {x + 1}{\sqrt{3}} \right)^2 + 1}

    write the integral in that form and then do a substitution of u = \frac {x + 1}{\sqrt{3}}. you should be able to handle it from there
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by twilightstr View Post
    integrate (x-3)/(x^2 + 2x+4).
    \frac{x-3}{x^2+2x+4}=\frac{x-3}{(x+1)^2+3}

    so try the substitution: u=x+1 to get:

    \frac{x-3}{x^2+2x+4}=\frac{x-3}{(x+1)^2+3}=\frac{u}{u^2+3}-\frac{4}{u^2+3}

    CB
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