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Thread: [SOLVED] Antiderivative

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Antiderivative

    $\displaystyle \int{\frac{x-1}{(x+1)^2(x^2+3)}}dx = \int{\left(\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C x + D}{x^2+3}\right)dx}$

    $\displaystyle A(x+1)^2(x^2+3) + B(x+1)(x^2+3) + (Cx + D)(x+1)^3 = x-1 \implies$


    $\displaystyle (A+C)x^4 + (2A+B+3C+D)x^3 + (4A+B+3C+3D)x^2 +
    (6A+3B+C+3D)x +$$\displaystyle (3A+3B+D) = x-1$

    Good so far? I fail to properly solve the equation system every time. I could really need another pair of eyes to look at what I've done so far, because this is driving me crazy.

    $\displaystyle x^4: A+C=0$
    $\displaystyle x^3: 2A+B+3C+D=0$
    $\displaystyle x^2: 4A+B+3C+3D =0$
    $\displaystyle x^1: 6A+3B+C+3D=1$
    $\displaystyle x^0: 3A+3B+D=-1$
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  2. #2
    MHF Contributor

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    Seeing the answer may help see how to do it.
    Attached Thumbnails Attached Thumbnails [SOLVED] Antiderivative-partfrac.gif  
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  3. #3
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    Quote Originally Posted by Spec View Post
    $\displaystyle \int{\frac{x-1}{(x+1)^2(x^2+3)}}dx = \int{\left(\frac{A}{x+1}+\frac{B}{(x+1)^2}+\frac{C x + D}{x^2+3}\right)dx}$

    $\displaystyle A(x+1)^2(x^2+3) + B(x+1)(x^2+3) + (Cx + D)(x+1)^3 = x-1 \implies$

    Mr F says: No. You've made a lot of mistakes. It should be:

    $\displaystyle {\color{red}A(x+1)(x^2+3) + B(x^2+3) + (Cx + D)(x+1)^2 = x-1}$.

    The simplest way is to then substitute convenient values of x into the above. For example, x = -1 will let you get B very easily.


    [snip]
    ..
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