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Thread: Solving an integral using partial fractions

  1. #1
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    Solving an integral using partial fractions

    I must solve using partial fractions:

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


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

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

    Then I did:

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

    I constructed the system:

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

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

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


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

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

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

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

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

    can be written as

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

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

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

    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:

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

    [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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