# How do you express as a partial fraction x/((x+1)(x^2 -1))?

Aug 22, 2015

$\frac{- \frac{1}{4}}{x + 1} + \frac{\frac{1}{2}}{x + 1} ^ 2 + \frac{\frac{1}{4}}{x - 1} = \frac{1}{4} \left[\frac{- 1}{x + 1} + \frac{2}{x + 1} ^ 2 + \frac{1}{x - 1}\right]$

#### Explanation:

First we need to finish factoring the denominator into factors that are irreducible using Real coefficients:

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

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

So we need:

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

Clear the denominator (multiply by $\left({\left(x + 1\right)}^{2} \left(x - 1\right)\right)$on both sides), to get:

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

$A {x}^{2} - A + B x - B + C {x}^{2} + 2 C x + C = x$

$A {x}^{2} + C {x}^{2} + B x + 2 C x - A - B + C = 0 {x}^{2} + 1 x + 0$

So we need to solve:

$A + C = 0$
$B + 2 C = 1$
$- A - B + C = 0$

From the first equation, we get: $A = - C$ and we can substitue in the third equation to get

Eq 2: $B + 2 C = 1$
and: $- B + 2 C = 0$

Adding gets us $C = \frac{1}{4}$, so $A = - \frac{1}{4}$ and subtracting gets us $B = \frac{1}{2}$

$\frac{A}{x + 1} + \frac{B}{x + 1} ^ 2 + \frac{C}{x - 1} = \frac{- \frac{1}{4}}{x + 1} + \frac{\frac{1}{2}}{x + 1} ^ 2 + \frac{\frac{1}{4}}{x - 1}$

$= \frac{1}{4} \left[\frac{- 1}{x + 1} + \frac{2}{x + 1} ^ 2 + \frac{1}{x - 1}\right]$