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Thread: Intergrate Calc dont you love it

  1. #1
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    Intergrate Calc dont you love it

    Intergrate (2x^2+4x+22) / (x^2+2x+10) dx
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  2. #2
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    Hello, Bust20001

    $\displaystyle \int\frac{2x^2+4x+22}{x^2+2x+10}\,dx$

    Divide: . $\displaystyle \int\left(2 + \frac{2}{x^2 +2x+10}\right)\,dx$

    . . . . . $\displaystyle =\;\;\int\left(2 + \frac{2}{(x+1)^2+9}\right)\,dx$

    . . . . . $\displaystyle = \;\;2x + \frac{2}{3}\arctan\left(\frac{x+1}{3}\right)+ C$

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  3. #3
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    Quote Originally Posted by Bust2000 View Post
    Intergrate (2x^2+4x+22) / (x^2+2x+10) dx
    $\displaystyle \frac{2x^2 + 4x + 22}{x^2 + 2x + 10} = 2\left( \frac{x^2 + 2x + 11}{x^2 + 2x + 10} \right) = 2\left( \frac{(x^2 + 2x + 10) + 1}{x^2 + 2x + 10} \right)$


    $\displaystyle = 2\left( 1 + \frac{1}{x^2 + 2x + 10} \right) = 2\left( 1 + \frac{1}{(x + 1)^2 + 9} \right) = \left( 1 + \frac{1}{3} \, \frac{3}{(x + 1)^2 + 3^2} \right)$.

    You should have better luck with this form .....
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  4. #4
    Junior Member roy_zhang's Avatar
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    Quote Originally Posted by Bust2000 View Post
    Intergrate (2x^2+4x+22) / (x^2+2x+10) dx
    When your integrand is in the form $\displaystyle \frac{P(x)}{Q(x)}$, where $\displaystyle P(x),Q(x)$ are both polynomials, and the order of $\displaystyle P(x)$ is greater or equal to the order of $\displaystyle Q(x)$; always try long division first before integrating.

    Here we have:
    $\displaystyle \frac{2x^2+4x+22}{x^2+2x+10} = 2+\frac{2}{x^2+2x+10}$

    Now,
    $\displaystyle \int\frac{2x^2+4x+22}{x^2+2x+10}dx = \int (2+\frac{2}{x^2+2x+10})dx = \int2dx + \int[\frac{2}{(x+1)^2+3^2}]dx$

    use $\displaystyle u$ substitution (let $\displaystyle u= x+1$)

    we got:

    $\displaystyle \int \frac{2x^2+4x+22}{x^2+2x+10}dx=2x + \frac{2}{3}\arctan(\frac{x+1}{3}) + C$
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