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Math Help - Solving an integral using partial fractions

  1. #1
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    Solving an integral using partial fractions

    I must solve using partial fractions:

    \displaystyle\int_{}^{}\displaystyle\frac{dx}{8x^3  +1}


    The only real root for the denominator: -\displaystyle\frac{1}{2}

    Then 8x^3+1=(x+\displaystyle\frac{1}{2})(8x^2-4x+2)

    Then I did:

    \displaystyle\frac{1}{(x+\displaystyle\frac{1}{2})  (8x^2-4x+2)}=\displaystyle\frac{A}{(x+\displaystyle\frac  {1}{2})}+\displaystyle\frac{Bx+C}{(8x^2-4x+2)}

    I constructed the system:

    A=\displaystyle\frac{1}{6} B=\displaystyle\frac{-4}{3} C=\displaystyle\frac{4}{3}

    \displaystyle\int_{}^{}\displaystyle\frac{dx}{8x^3  +1}=\displaystyle\frac{1}{6}\displaystyle\int_{}^{  }\displaystyle\frac{dx}{(x+\displaystyle\frac{1}{2  })}dx-\displaystyle\frac{4}{3}\displaystyle\int_{}^{}\di  splaystyle\frac{x-1}{(8x^2-4x+2)}dx

    Completing the square 8x^2-4x+2=8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{  2}


    \displaystyle\int_{}^{}\displaystyle\frac{x-1}{(8x^2-4x+2)}dx=\displaystyle\int_{}^{}\displaystyle\frac  {x-1}{8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{  2}}dx

    \displaystyle\int_{}^{}\displaystyle\frac{x-1}{8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{  2}}=\displaystyle\int_{}^{}\displaystyle\frac{x}{8  (x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{  2}}dx-\displaystyle\int_{}^{}\displaystyle\frac{1}{8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{  2}}dx

    t=x-\displaystyle\frac{1}{4}\Rightarrow{x=t+\displayst  yle\frac{1}{4}}
    dt=dx

    \displaystyle\int_{}^{}\displaystyle\frac{x}{8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{  2}}dx=\displaystyle\int_{}^{}\displaystyle\frac{t+  \displaystyle\frac{1}{4}}{8t^2+\displaystyle\frac{  3}{2}}dt=\displaystyle\frac{1}{8}\displaystyle\int  _{}^{}\displaystyle\frac{t}{t^2+\displaystyle\frac  {3}{2}}dt+\displaystyle\frac{1}{32}\displaystyle\i  nt_{}^{}\displaystyle\frac{dt}{t^2+\displaystyle\f  rac{3}{2}}

    Am I on the right way? Its too tedious. The statement says I must use partial fractions, but anyway if you see a simpler way of solving it let me know.
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  2. #2
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    \frac{4}{3}\int \frac{x-1}{(8x^2 - 4x + 2)} dx

    can be written as

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

    The numerator can be written as 1/8(8x - 2 - 6)

    \frac{2}{3\times 8}[\int \frac{8x-2}{(4x^2 - 2x + 1)}dx - \int \frac{6}{(4x^2 - 2x + 1)}dx][

    The first integration is in form of f'(x)/f(x), which is equal to log[f(x)] and second integration can be solved by complting the square in the dinominator.
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  3. #3
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    Quote Originally Posted by Ulysses View Post
    I must solve using partial fractions:

    \displaystyle\int_{}^{}\displaystyle\frac{dx}{8x^3  +1}

    [snip]

    Am I on the right way? Its too tedious. The statement says I must use partial fractions, but anyway if you see a simpler way of solving it let me know.
    There is no avoiding tedious algebra.

    See here: integrate 1/(8x^3 + 1) - Wolfram|Alpha

    and click on Show steps.
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