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What is the name of this method?

  1. Sep 23, 2012 #1

    1. The problem statement, all variables and given/known data
    Integrate:
    [tex]\int \frac{1}{x^3 - 27} dx[/tex]

    2. Relevant equations
    [tex]\int \frac{1}{x^3 - 27} dx[/tex]
    [tex]\frac{1}{x^3 - 27} = \frac{1}{(x - 3)(x^2 + 3x + 9)} = \frac{A}{x - 3} + \frac{Bx + C}{x^2 + 3x + 9}[/tex]
    [tex]1 = A(x^2 + 3x + 9) + (Bx + C)(x - 3)[/tex]

    3. The attempt at a solution

    The first step involves factoring and splitting the denominator with some method:
    [tex]\frac{1}{x^3 - 27} = \frac{1}{(x - 3)(x^2 + 3x + 9)} = \frac{A}{x - 3} + \frac{Bx + C}{x^2 + 3x + 9}[/tex]
    What is the name of this method?

    The next step involves normalizing the equation:
    [tex]1 = A(x^2 + 3x + 9) + (Bx + C)(x - 3)[/tex]
    How was the coefficients of B and C derived and solved?

    The solution only states that:
    [tex]A = \frac{1}{27}[/tex]
    [tex]0 = \frac{1}{27} + B[/tex]
    [tex]1 = \frac{1}{3} + 3C[/tex]
    And:
    [itex]B = -\frac{1}{27}[/itex] and [itex]C = -\frac{2}{9}[/itex]

    I can see how the [itex]A[/itex] coefficient was derived from the x-axis zero intercept, however, it is not entirely clear to me how these other coefficients were derived and solved?

    Reference:
    Derivation of equation and solution
     
    Last edited: Sep 23, 2012
  2. jcsd
  3. Sep 23, 2012 #2

    CAF123

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    Gold Member

    This is the method of partial fractions to reduce [tex] \int \frac{1}{x^3 -27} dx [/tex] into something which can be integrated.

    To find A,B and C you will need to solve simultaneous equations. To easily solve for A, let x =3 and you should get the required A = 1/27. To find B and C, let x = 0 to get one eqn in B and C and let x = 1 to get another. Two eqns, two unknowns (B and C) - you can solve this. The choice of x here is arbritary (except x =3 since the (Bx +C) term will vanish).
     
    Last edited: Sep 23, 2012
  4. Sep 23, 2012 #3

    HallsofIvy

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    Staff Emeritus
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    Actually, because the last term in involved "Bx- C", and you have already found A, setting x= 0 will give an equation in C only.
     
  5. Sep 23, 2012 #4

    CAF123

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    Gold Member

    Yes, I overlooked this. If you have something like [tex] \frac{A}{x} + \frac{(Bx +c)}{x^2 +bx +c}, [/tex] then you will probably have to use simultaneous eqns.
     
  6. Sep 23, 2012 #5

    Solving the coefficients:
    Given:
    [tex]1 = A(x^2 + 3x + 9) + (Bx + C)(x - 3)[/tex]

    Set x = 3 to zero out the [itex](Bx + C)[/itex] term and numerically integrate via substitution:
    [tex]1 = A(3^2 + 3(3) + 9) + (B(3) + C)(3 - 3) = A(27)[/tex]
    [tex]1 = A(27)[/tex]
    Solve for [itex]A[/itex]:
    [tex]\boxed{A = \frac{1}{27}}[/tex]
    Set x = 0 to zero out the [itex]Bx[/itex] term and numerically integrate via substitution:
    [tex]1 = \frac{(0^2 + 3(0) + 9)}{27} + (B(0) + C)(0 - 3) = \frac{9}{27} - 3C[/tex]
    [tex]1 = \frac{9}{27} - 3C[/tex]
    [tex]-\frac{27}{27} + \frac{9}{27} = -\frac{18}{27} = -\frac{2}{3} = 3C[/tex]
    Solve for [itex]C[/itex]:
    [tex]\boxed{C = -\frac{2}{9}}[/tex]
    Set x = 1 as arbitrary and numerically integrate via substitution:
    [tex]1 = \frac{(1^2 + 3(1) + 9)}{27} + \left(B(1) - \frac{2}{9} \right)(1 - 3) = \frac{13}{27} - 2B + \frac{3}{3} \times \frac{4}{9} = \frac{13}{27} - 2B + \frac{12}{27} = \frac{25}{27} - 2B[/tex]
    [tex]1 = \frac{25}{27} - 2B[/tex]
    [tex]-\frac{27}{27} + \frac{25}{27} = -\frac{2}{27} = 2B[/tex]
    Solve for [itex]B[/itex]:
    [tex]\boxed{B = -\frac{1}{27}}[/tex]
     
    Last edited: Sep 23, 2012
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