(adsbygoogle = window.adsbygoogle || []).push({});

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

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: What is the name of this method?

**Physics Forums | Science Articles, Homework Help, Discussion**