Homework Help: What is the name of this method?

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

 

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]