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Math Help - integrate rational functions

  1. #1
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    integrate rational functions

    Hi, im trying to integrate this function: (3x^2-10)/(x^2-4x+4)dx. I got it down to int 3+[(12x-22)/((x-2)^2 by long division. Then i get [A/(x-2)]+[B/(x-2)^2], i get stuck on trying to figure out what A and B are equal to. Can somone please help me with this?
    thank you
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  2. #2
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    Quote Originally Posted by cowboys111 View Post
    Hi, im trying to integrate this function: (3x^2-10)/(x^2-4x+4)dx. I got it down to int 3+[(12x-22)/((x-2)^2 by long division. Then i get [A/(x-2)]+[B/(x-2)^2], i get stuck on trying to figure out what A and B are equal to. Can somone please help me with this?
    thank you
    Get everything over the common denominator of (x - 2)^2 and equate the numerators of each side:

    12x - 22 = A(x - 2) + B

    => 12x - 22 = Ax + (B - 2A)

    => A = 12 and -22 + B - 2A .......
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  3. #3
    Super Member PaulRS's Avatar
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    We want: <br />
\int {\frac{{3 \cdot x^2  - 10}}<br />
{{x^2  - 4x + 4}}dx} <br />

    Note that <br />
3x^2  - 10 = 3 \cdot \left( {x^2  - 4} \right) + 2 = 3 \cdot \left( {x - 2} \right) \cdot \left( {x + 2} \right) + 2<br />
and <br />
x^2  - 4x + 4 = \left( {x - 2} \right)^2 <br />

    Thus: <br />
\int {\frac{{3x^2  - 10}}<br />
{{\left( {x - 2} \right)^2 }}} dx = \int {\frac{{3 \cdot \left( {x - 2} \right) \cdot \left( {x + 2} \right) + 2}}<br />
{{\left( {x - 2} \right)^2 }}} dx = 3 \cdot \int {\frac{{x + 2}}<br />
{{x - 2}}} dx + 2 \cdot \int {\frac{{dx}}<br />
{{\left( {x - 2} \right)^2 }}} dx<br />

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