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

Dec 20, 2016

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

#### Explanation:

Let's do the decomposition into partial fractions

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

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

Therefore,

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

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

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

Coefficients of ${x}^{3}$, $\implies$, $0 = C + D$, $\implies$, $C = - 3$

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

$- 1 = 2 + 2 B + 6 - 3$

$2 B = - 1 - 2 - 3 = - 6$, $\implies$, $B = - 3$

Therefore,

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