# How do you express (2x - 1) / [(x - 1)^3 (x - 2)] in partial fractions?

Jul 14, 2017

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

#### Explanation:

Let's perform the decomposition into partial fractions

$\frac{2 x - 1}{{\left(x - 1\right)}^{3} \left(x - 2\right)} = \frac{A}{x - 1} ^ 3 + \frac{B}{x - 1} ^ 2 + \frac{C}{x - 1} + \frac{D}{x - 2}$

$= \frac{A \left(x - 2\right) + B \left(x - 1\right) \left(x - 2\right) + C {\left(x - 1\right)}^{2} \left(x - 2\right) + D {\left(x - 1\right)}^{3}}{{\left(x - 1\right)}^{3} \left(x - 2\right)}$

The denominators are the same, we compare the numerators

2x-1=A(x-2)+B(x-1)(x-2)+C(x-1)^2(x-2))+D(x-1)^3

Let $x = 1$, $\implies$, $1 = - A$, $\implies$, $A = - 1$

Let $x = 2$, $\implies$, $3 = D$

Coefficients of ${x}^{3}$

$0 = C + D$, $\implies$, $C = - D = - 3$

Let $x = 0$, $- 1 = - 2 A + 2 B - 2 C - D$

$2 B = 2 A + 2 C + D - 1 = - 2 - 6 + 3 - 1 = - 6$

$B = - 3$

Therefore,

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