# What is the name of this method?

1. Sep 23, 2012

### Orion1

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

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

3. The attempt at a solution

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

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

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

I can see how the $A$ 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. Sep 23, 2012

### CAF123

This is the method of partial fractions to reduce $$\int \frac{1}{x^3 -27} dx$$ 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
3. Sep 23, 2012

### HallsofIvy

Staff Emeritus
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.

4. Sep 23, 2012

### CAF123

Yes, I overlooked this. If you have something like $$\frac{A}{x} + \frac{(Bx +c)}{x^2 +bx +c},$$ then you will probably have to use simultaneous eqns.

5. Sep 23, 2012

### Orion1

Solving the coefficients:
Given:
$$1 = A(x^2 + 3x + 9) + (Bx + C)(x - 3)$$

Set x = 3 to zero out the $(Bx + C)$ term and numerically integrate via substitution:
$$1 = A(3^2 + 3(3) + 9) + (B(3) + C)(3 - 3) = A(27)$$
$$1 = A(27)$$
Solve for $A$:
$$\boxed{A = \frac{1}{27}}$$
Set x = 0 to zero out the $Bx$ term and numerically integrate via substitution:
$$1 = \frac{(0^2 + 3(0) + 9)}{27} + (B(0) + C)(0 - 3) = \frac{9}{27} - 3C$$
$$1 = \frac{9}{27} - 3C$$
$$-\frac{27}{27} + \frac{9}{27} = -\frac{18}{27} = -\frac{2}{3} = 3C$$
Solve for $C$:
$$\boxed{C = -\frac{2}{9}}$$
Set x = 1 as arbitrary and numerically integrate via substitution:
$$1 = \frac{(1^2 + 3(1) + 9)}{27} + \left(B(1) - \frac{2}{9} \right)(1 - 3) = \frac{13}{27} - 2B + \frac{3}{3} \times \frac{4}{9} = \frac{13}{27} - 2B + \frac{12}{27} = \frac{25}{27} - 2B$$
$$1 = \frac{25}{27} - 2B$$
$$-\frac{27}{27} + \frac{25}{27} = -\frac{2}{27} = 2B$$
Solve for $B$:
$$\boxed{B = -\frac{1}{27}}$$

Last edited: Sep 23, 2012