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

Feb 16, 2017

The answer is $= \frac{- \frac{1}{4}}{x + 1} + \frac{\frac{1}{2}}{x - 1} ^ 2 + \frac{\frac{1}{4}}{x - 1}$

#### Explanation:

Let's factorise the denominator

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

Therefore,

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

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

Let's perform the decomposition into partial fractions

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

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

The denominators are the same, we compare the numerators

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

Let $x = - 1$, $\implies$, $- 1 = 4 A$, $\implies$, $A = - \frac{1}{4}$

Let $x = 1$, $\implies$, $1 = 2 B$, $\implies$, $B = \frac{1}{2}$

Coefficients of ${x}^{2}$

$0 = A + C$, $\implies$, $C = - A = \frac{1}{4}$

Therefore,

$\frac{x}{\left({x}^{2} - 1\right) \left(x - 1\right)} = \frac{- \frac{1}{4}}{x + 1} + \frac{\frac{1}{2}}{x - 1} ^ 2 + \frac{\frac{1}{4}}{x - 1}$