# How do you use partial fraction decomposition to decompose the fraction to integrate 2/(x^3-x^2)?

Aug 3, 2015

$\frac{2}{{x}^{3} - {x}^{2}} = - \frac{2}{x} - \frac{2}{x} ^ 2 + \frac{2}{x - 1}$

#### Explanation:

Factor the denominator:

${x}^{3} - {x}^{2} = {x}^{2} \left(x - 1\right)$

So we need, $A , B , \text{ and } C$ to make:

$\frac{A}{x} + \frac{B}{x} ^ 2 + \frac{C}{x - 1} = \frac{2}{{x}^{2} \left(x - 1\right)}$

$\frac{A x \left(x - 1\right) + B \left(x - 1\right) + C {x}^{2}}{{x}^{2} \left(x - 1\right)} = \frac{2}{{x}^{2} \left(x - 1\right)}$

$A {x}^{2} - A x + B x - B + C {x}^{2} = 2$

$\left(A + C\right) {x}^{2} + \left(- A + B\right) x + \left(- B\right) = 0 {x}^{2} + 0 x + 2$

So

$A + C = 0$
$- A + B = 0$
$- B = 2$

And we find:

$B = - 2 ,$ $\text{ "A=-2", }$ and $\text{ } C = 2$

$\frac{2}{{x}^{3} - {x}^{2}} = - \frac{2}{x} - \frac{2}{x} ^ 2 + \frac{2}{x - 1}$