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Math Help - [SOLVED] Partial fraction in an integral

  1. #1
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    [SOLVED] Partial fraction in an integral

    \int_{1}^{2} \frac {4y^2-7y-12}{y(y+2)(y-3)} ~dx

    So far I have \frac {A}{y} + \frac {B}{(y+2)} + \frac {C}{(y-3)}

    You have to combine them right and get a common denominator? So I have 4y^2-7y-12 = A(y+2)(y-3) + B(y)(y-3) + C(y)(y+2)

    But what do I do from here? So far, we have only done examples in class with 2 variables, A and B, but not with C. So its throwing me off.
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  2. #2
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    I'm assuming that the integral is: \int_{1}^{2} \frac {4y^2-7y-12}{y(y+2)(y-3)} ~d{\color{red}y}

    You've done well. Just multiply the right-hand side a bit more and collect like terms:
    4y^2-7y-12 = A(y+2)(y-3) + B(y)(y-3) + C(y)(y+2)

    4y^2-7y-12 = Ay^2 - Ay - 6A \: \: +  \: \: By^2 - 3By \: \: + \: \: Cy^2+2Cy

    4y^2-7y-12 = (A + B + C)y^2 + (-A - 3B + 2C)y - 6A

    Equating the coefficients, we see that:
    A + B + C = 4 \qquad -A - 3B + 2C = -7 \qquad -6A = -12

    etc.etc.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by redman223 View Post
    \int_{1}^{2} \frac {4y^2-7y-12}{y(y+2)(y-3)} ~d{\color{red}y}

    So far I have \frac {A}{y} + \frac {B}{(y+2)} + \frac {C}{(y-3)}

    You have to combine them right and get a common denominator? So I have 4y^2-7y-12 = A(y+2)(y-3) + B(y)(y-3) + C(y)(y+2).......................................(1)

    But what do I do from here? So far, we have only done examples in class with 2 variables, A and B, but not with C. So its throwing me off.
    Alternatively.

    call your equation (1):

    now, plug in y = 0 into (1). this will get rid of B and C and give you an equation to solve for A

    then, plug in y = 3 into (1). this will get rid of A and B and give you an equation with C as the only unknown, so you can solve for it

    then, plug in y = -2 into (1). this will get rid of A and C and...you know the drill
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