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Homework Help: Integral using partial fraction

  1. Jul 1, 2010 #1
    1. The problem statement, all variables and given/known data

    I must solve using partial fractions:

    [tex]\displaystyle\int_{}^{}\displaystyle\frac{dx}{8x^3+1}[/tex]

    3. The attempt at a solution

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

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

    Then I did:

    [tex]\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)}[/tex]

    I constructed the system:

    [tex]A=\displaystyle\frac{1}{6}[/tex] [tex]B=\displaystyle\frac{-4}{3}[/tex] [tex]C=\displaystyle\frac{4}{3}[/tex]

    [tex]\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_{}^{}\displaystyle\frac{x-1}{(8x^2-4x+2)}dx[/tex]

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


    [tex]\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[/tex]

    [tex]\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[/tex]

    [tex]t=x-\displaystyle\frac{1}{4}\Rightarrow{x=t+\displaystyle\frac{1}{4}}[/tex]
    [tex]dt=dx[/tex]

    [tex]\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\int_{}^{}\displaystyle\frac{dt}{t^2+\displaystyle\frac{3}{2}}[/tex]

    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.
     
    Last edited: Jul 1, 2010
  2. jcsd
  3. Jul 1, 2010 #2

    Mark44

    Staff: Mentor

    This is the correct approach. You might have made life a little easier on yourself by factoring 8x^3 + 1 into (2x + 1)(4x^2 -2x + 1), thereby eliminating at least some of the fractions.
     
  4. Jul 1, 2010 #3
    Didn't realize of it. Thanks Mark44. Too many fractions really. Its killing me :P
     
  5. Jul 1, 2010 #4
    This is the kind of problem that you shouls do once, for whatever reason (!), and then use a computer for afterward! Good luck. wolframalpha can verify your results, in case you weren't aware of that resource.
     
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