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Math Help - Partial Fractions

  1. #1
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    Partial Fractions

    I'm not certain why but I can't seem to get the right answer no matter how many times I run through this

    Integrate (x^2) / ((x - 3)(x + 2)^2)


    (A / x - 3) + (B / x + 2) + (C / (x + 2)^2)

    My values for A and C seem correct from the method I was taught by the professor

    A = 9 / 25 C = - 4 / 5

    Now to get B, I equated for cubics.

    x^2 = ((9/25)(x + 2)^3) + B(x - 3)(x + 2)^2 - (4/5)(x - 3)(x + 2)

    Now the B I got was - 9 / 25.

    Can someone tell me if I've done this right so far? Thanks
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  2. #2
    Eater of Worlds
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    You didn't multiply by (x-3)(x+2)(x+2)^{2} did you?.

    Should use (x-3)(x+2)^{2}

    (x+2)^{2}(x-3)\cdot\frac{A}{x-3}+(x+2)^{2}(x-3)\cdot\frac{B}{x+2}+(x+2)^{2}(x-3)\cdot\frac{C}{(x+2)^{2}}

    Cancel appropriately, expand out, and we get:

    Ax^{2}+4Ax+4A+Bx^{2}-Bx-6B+Cx-3C=x^{2}

    Equate coefficients:

    A+B=1

    4A-B+C=0

    4A-6B-3C=0

    Solve the system and we get:

    A=\frac{9}{25}, \;\ B=\frac{16}{25}, \;\ C=\frac{-4}{5}

    So, we get:

    \frac{9}{25(x-3)}+\frac{16}{25(x+2)}-\frac{4}{5(x+2)^{2}}
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  3. #3
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    Yea I did do that . Thanks for the help I appreciate it.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    I'm not certain why but I can't seem to get the right answer no matter how many times I run through this

    Integrate (x^2) / ((x - 3)(x + 2)^2)


    (A / x - 3) + (B / x + 2) + (C / (x + 2)^2)

    My values for A and C seem correct from the method I was taught by the professor

    A = 9 / 25 C = - 4 / 5

    Now to get B, I equated for cubics.

    x^2 = ((9/25)(x + 2)^3) + B(x - 3)(x + 2)^2 - (4/5)(x - 3)(x + 2)

    Now the B I got was - 9 / 25.

    Can someone tell me if I've done this right so far? Thanks
    Alternatively,

    \frac {x^2}{(x - 3)(x + 2)^2} = \frac A{x - 3} + \frac B{x + 2} + \frac C{(x + 2)^2}

    multiply through by (x - 3)(x + 2)^2

    \Rightarrow x^2 = A(x + 2)^2 + B(x - 3)(x + 2) + C(x - 3)

    Let x = -2
    \Rightarrow 4 = -5C \implies \boxed{C = - \frac 45}

    Let x = 3
    \Rightarrow 9 = 25A \implies \boxed{A = \frac 9{25}}

    Let x = 0 ....just some random value, 0 is easy to work with:
    \Rightarrow 0 = 4A - 6B - 3C

    plug in the values for A and C and solve for B, we get: \boxed{B = \frac {16}{25}}. as the great galactus said
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  5. #5
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    Thanks Jhevon for the alternative method!! I'll remember that
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