# How do you write the partial fraction decomposition of the rational expression  5/(x^2-1)^2?

Nov 1, 2016

The answer is $= \frac{5}{4 {\left(x + 1\right)}^{2}} + \frac{5}{4 \left(x + 1\right)} + \frac{5}{4 {\left(x - 1\right)}^{2}} - \frac{5}{4 \left(x - 1\right)}$

#### Explanation:

Let's factorise the denominator (x^2-1)^2=(x+1)(x-1(x+1)(x-1)
So $\frac{5}{{x}^{2} - 1} ^ 2 = \frac{5}{{\left(x + 1\right)}^{2} {\left(x - 1\right)}^{2}}$
$= \frac{A}{x + 1} ^ 2 + \frac{B}{x + 1} + \frac{C}{x - 1} ^ 2 + \frac{D}{x - 1}$
$5 =$$A {\left(x - 1\right)}^{2} + B \left(x + 1\right) {\left(x - 1\right)}^{2} + C {\left(x + 1\right)}^{2} + D \left(x - 1\right) {\left(x + 1\right)}^{2}$
Let $x = 1$ then $5 = 4 C$$\implies$$C = \frac{5}{4}$
$x = - 1$ then $5 = 4 A$$\implies$$A = \frac{5}{4}$
$x = 0$$\implies$ $5 = A + B + C - D$
$B - D = 5 - \frac{5}{4} - \frac{5}{4} = \frac{5}{2}$
Coefficients of ${x}^{3}$ then $0 = B + D$
$B = \frac{5}{4}$ and $D = - \frac{5}{4}$
So $= \frac{A}{x + 1} ^ 2 + \frac{B}{x + 1} + \frac{C}{x - 1} ^ 2 + \frac{D}{x - 1}$
$= \frac{5}{4 {\left(x + 1\right)}^{2}} + \frac{5}{4 \left(x + 1\right)} + \frac{5}{4 {\left(x - 1\right)}^{2}} - \frac{5}{4 \left(x - 1\right)}$