- #1
Orion1
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Homework Statement
Integrate:
[tex]\int \frac{1}{x^3 - 27} dx[/tex]
Homework 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]
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:
http://www.scienceforums.net/index.php?app=core&module=attach§ion=attach&attach_id=4008
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