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Math Help - Partial fractions (integration)

  1. #1
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    Partial fractions (integration)

    Evaluate the integral:


    \int \frac{1}{x^3 - 1} dx

    I got to the final step:

    =ln|3x - 1| - \int \frac{3x}{x^2 + x + 1}dx + \int \frac{\frac{2}{3}}{x^2 + x + 1} dx

    but dont know how to integrate the 2 integrals due to the irreducible quadratic factors of the denominator.
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  2. #2
    Senior Member eumyang's Avatar
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    What did you get for your partial-fraction decomposition? Because I got
    \dfrac{1/3}{x - 1} - \dfrac{(1/3)x + 2/3}{x^2 + x + 1}
    \dfrac{1}{3(x - 1)} - \dfrac{x + 2}{3(x^2 + x + 1)}
    but it doesn't look like it would lead to your final step.
    Last edited by eumyang; August 9th 2010 at 11:00 AM.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    The procedure of integrate the above is quite standard. I don't like to use wolfram in many occasions, but this is an exception...
    int 1/(X^3-1) - Wolfram|Alpha

    (Press on 'show steps')
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  4. #4
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    Quote Originally Posted by SyNtHeSiS View Post
    Evaluate the integral:


    \int \frac{1}{x^3 - 1} dx

    I got to the final step:

    =ln|3x - 1| - \int \frac{3x}{x^2 + x + 1}dx + \int \frac{\frac{2}{3}}{x^2 + x + 1} dx

    but dont know how to integrate the 2 integrals due to the irreducible quadratic factors of the denominator.
    When I split your original integral into partial fractions, I get

    \displaystyle\huge\frac{1}{x^3-1}=\frac{1}{(x-1)\left(x^2+x+1\right)}

    since

    (x-1)\left(x^2+bx+c\right)=x^3-1

    x^3+bx^2+cx-x^2-bx-c=x^3-1

    b=1,\ c=1

    Then

    \displaystyle\huge\frac{1}{(x-1)\left(x^2+x+1\right)}=\frac{A}{x-1}+\frac{B+Cx}{x^2+x+1}

    giving

    A=\frac{1}{3},\ B=-\frac{2}{3},\ C=-\frac{1}{3}
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  5. #5
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by SyNtHeSiS View Post
    Evaluate the integral:


    \int \frac{1}{x^3 - 1} dx

    I got to the final step:

    =ln|3x - 1| - \int \frac{3x}{x^2 + x + 1}dx + \int \frac{\frac{2}{3}}{x^2 + x + 1} dx

    but dont know how to integrate the 2 integrals due to the irreducible quadratic factors of the denominator.
    We should note that

    \int \frac{1}{x^3 - 1} dx = \int \frac{1}{x^3 - 1^3} dx

    And,

     x^3 - 1^3 = (x-1)(x^2 +x + 1)

    Resulting in,

     \int \frac{1}{ (x-1)(x^2 +x + 1) } dx

    Partial Fractions yields

     \frac{ A }{x-1} + \frac{ Bx + C }{ (x^2 +x + 1) }

     \frac{ A (x^2 +x + 1) + B(x^2 - x) + C(x-1) } { (x-1)(x^2 +x + 1) }

     1: A - C = 1

     x: A -B + C = 0

     x^2: A + B = 0

    This yields,

     C = - \frac{2}{3}

     A = \frac{1}{3}

     B = - \frac{1}{3}

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

    The second integral can be evaluated as follows,

     \frac{1}{3} \int \frac{x + 2}{(x^2 +x + 1) }dx

     \frac{1}{3} \int \frac{x}{ (x^2 +x + 1) }dx + \frac{2}{3} \int \frac{1}{(x^2 +x + 1) }dx

    Now let's look at

     \frac{2}{3} \int \frac{1}{(x^2 +x + 1) }dx

    What happens when we try to make this look like the inverse tan integral?

     \frac{2}{3} \int \frac{4}{(2x+1)^2 + ( \sqrt{3} )^2 }dx

    If you expand inside the integral (to check that we have done this correctly) we will see that

     \frac{4}{(2x+1)^2 + ( \sqrt{3} )^2 } = \frac{4}{ 4x^2 + 4x + 4 } = \frac{1}{(x^2 +x + 1) }

    It should be obvious then that,

     \frac{2}{3} \int \frac{1}{(x^2 +x + 1) }dx = \frac{8}{3} \int \frac{1}{(2x+1)^2 + ( \sqrt{3} )^2 }dx = \frac{8}{3 \sqrt{3} } tan^{-1} ( \frac{ 2x+1}{ \sqrt{3} }) + C

    I leave you to consider

     \frac{2}{3} \int \frac{x}{(x^2 +x + 1) }dx
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