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Math Help - integration

  1. #1
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    integration

    how do u do this integration? ive tried so many times and it just doesnt work!! i tried the completing the square way but i didnt get the right answer.

    integral xdx/(x^2-2x+3)
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  2. #2
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    Notice that \int \frac{x}{x^2-2x+3} = \int \frac{x}{(x-1)^2+2}

    Sub u = x-1 and you'll get

    \int \frac{u+1}{u^2+2}

    Splitting it will get you two simpler integrals that I think you should see right away.
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  3. #3
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    Quote Originally Posted by chukie View Post
    how do u do this integration? ive tried so many times and it just doesnt work!! i tried the completing the square way but i didnt get the right answer.

    integral xdx/(x^2-2x+3)
    \frac{x}{x^2-2x+3} = \frac{x-1}{x^2-2x+3} + \frac{1}{x^2-2x+3}

    Now, \int \frac{x-1}{x^2-2x+3} dx = \frac{1}{2}\int \frac{(x^2-2x+3)'}{x^2-2x+3}dx (Think logarithm here).

    And, \int \frac{dx}{x^2 - 2x+3} = \int \frac{dx}{(x-1)^2 + (\sqrt{2})^2} dx (Think arctangent).
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