# How do you use partial fraction decomposition to decompose the fraction to integrate (x^2-x+2)/(x(x-1)^2)?

Jul 27, 2015

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

#### Explanation:

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

Multiply both sides by $x {\left(x - 1\right)}^{2}$ (or get a common denominator and compare the numerators -- you'll get to the same place, but one approach may be clearer to you)

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

$A {x}^{2} - 2 A x + 1 + B {x}^{2} - B x + C x = {x}^{2} - x + 2$

$\left[A + B\right] {x}^{2} + \left[- 2 A - B + C\right] x + \left[A\right] = 1 {x}^{2} - 1 x + 2$

So we get:

$A + B = 1$
$- 2 A + B + C = - 1$
$A = 2$

Eq 1 and 3 get us $A = 2$ and $B = - 1$

Substituting in Eq 2, we get $C = 2$