- #1

- 970

- 3

## 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:

Derivation of equation and solution

Last edited: