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Thread: partial fractions

  1. #1
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    partial fractions

    Hia, I tried using partial fractions to integrate this expression but the integrator function on wolfram mathematics gave me a different answer.

    $\displaystyle \int \frac{x^3}{(x^2-9)(x+1)}dx$ $\displaystyle =\int \frac{x^3}{(x+3)(x-3)(x+1)}dx$


    integrals.wolfram.com gave me $\displaystyle \frac{1}{8}(8x+9ln(x-3)+ln(x+1)-18ln(x+3))$ where I presume the x is another constant

    whilst I get $\displaystyle \frac{7}{4}ln(x+3)+\frac{9}{8}ln(x-3)+\frac{1}{8}ln(x+1)$ which is almost there,,,but if anyone could help me that would be great! thankyou
    For my partial fractions I substituted x=-1,x=3 then when I did x=1 to calculate the coefficient above the term x+3 I kept getting 7/4. It must be a minor mistake which I can't see anywhere!!!!!
    $\displaystyle \frac{x^3}{(x+1)(x-3)(x+3)}=\frac{A}{x+3}+\frac{B}{x-3}+\frac{C}{x+1}$
    $\displaystyle x^3=A(X-3)(X+1)+B(X+3)(X+1)+C(x+3)(x-3)$
    x=1$\displaystyle \Longrightarrow$$\displaystyle C=\frac{1}{8}$X=3$\displaystyle \Longrightarrow$$\displaystyle B=\frac{27}{24}$X=1$\displaystyle \Longrightarrow$$\displaystyle A=\frac{7}{4}$???????
    Last edited by beaux; Nov 6th 2008 at 03:49 PM.
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  2. #2
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    Hello, beaux!

    You overlooked something quite important. . .


    $\displaystyle \int \frac{x^3}{(x+3)(x-3)(x+1)}dx$

    The degree of of the numerator is not less than the degree of the denominator.
    . . The fraction is "improper" (top-heavy).


    We must use Long Division . . .

    $\displaystyle x^3 \div(x^3 + x^2-9x-9) \;=\;1 - \underbrace{\frac{x^2-9x-9}{(x-3)(x+3)(x+1)}} $

    . . . . . .$\displaystyle \text{and apply Partial Fractions to }\overbrace{\text{this}}^{\uparrow}\text{ fraction.}$

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  3. #3
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    hia again,i dont understand?! Both the numerator and denominator are cubic but the denominator also has a linear expression with x also. Does this not make the denominator's degree bigger than the numerators?ie proper?
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