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

Dec 20, 2016

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

#### Explanation:

Let's rewrite the expression

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

The decomposition into partial fractions is

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

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

Therefore,

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

Let $x = 0$, $\implies$, 1=A#

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

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

Coefficients of ${x}^{4}$, $\implies$,$0 = A + C + E$, $\implies$, $C + E = - 1$

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

$C = E = - \frac{1}{2}$

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