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Thread: Integration partial fraction

  1. #1
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    Exclamation Integration partial fraction

    Integrate 1/x^2/(x^2+4)

    I tried:
    A/x + B/(x+1) + (C*x+D)/(x^2+4)

    1 = A(x^3+x^2+4*x+4)+B(x^3+4*x)+(C*x+D)*(x^2+x)

    1=(A+B+C)*x^3+(A+C+D)*x^2+(4*(A+B)+D)*x+4*A

    let x=0..... --> A=1/4
    (4*(A+B)+D)=0--> D=-4*B-1 --> D=-1/5
    (A+C+D)=0 --> (1/4)+C+-4*B-1=0 --> C=4*B+3/4 --> C= -1/20
    (A+B+C)=0 --> (1/4+B+(4*B+3/4) --> B=-1/5

    Then I plugged in:
    (1/4)/x + (-1/5)/(x+1) + ((-1/20)*x+(-1/20))/(x^2+4)

    I still don't know how to get to the solution of the partial fraction step which is supposed to be:
    -1/(4*(x^2+4))+1/(4*x^2)

    Help me out, I have posted this before and still didn't get it.
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  2. #2
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    Hello, LightEight!

    Your set-up is wrong . . .


    Integrate: .$\displaystyle \int \frac{dx}{x^2(x^2+4)}$

    You want: .$\displaystyle \frac{1}{x^2(x^2+4)} \;=\;\frac{A}{x} + {\color{red}\frac{B}{x^2}} + \frac{Cx+D}{x^2+4}$

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  3. #3
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    $\displaystyle \int (\frac{1}{x^2(x^2+4)})dx = \frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+4}$

    $\displaystyle 1=A(x(x^2+4))+B(x^2+4)+(Cx+D)x^2$

    $\displaystyle 1=Ax^3+4Ax+Bx^2+4B+Cx^3+Dx^2$

    $\displaystyle 1=(A+C)x^3+(B+D)x^2+4Ax+4B$
    $\displaystyle x=0$ ➔ $\displaystyle 1=4B$ ➔ $\displaystyle B=\frac{1}{4}$

    $\displaystyle (B+D)=0$ ➔ $\displaystyle (\frac{1}{4}+D)=0$ ➔ $\displaystyle D=-\frac{1}{4}$

    $\displaystyle (A+C)=0$ ➔ $\displaystyle A=-C$ I guess I can set $\displaystyle A=0$ and $\displaystyle C=0$

    $\displaystyle \frac{0}{x}+\frac{1}{4x^2}-\frac{1}{4x^2+16}$ ➔ $\displaystyle -\frac{1}{4x^2+16}+\frac{1}{4x^2}$
    Last edited by LightEight; Nov 10th 2009 at 02:01 PM.
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