# How do you express x/(16x^4-1) in partial fractions?

Nov 20, 2016

The answer is $= \frac{\frac{1}{8}}{2 x - 1} + \frac{\frac{1}{8}}{2 x + 1} + \frac{- \frac{x}{2}}{4 {x}^{2} + 1}$

#### Explanation:

Let's factorise the denominator

$16 {x}^{4} - 1 = \left(4 {x}^{2} - 1\right) \left(4 {x}^{2} + 1\right) = \left(2 x - 1\right) \left(2 x + 1\right) \left(4 {x}^{2} + 1\right)$

We can start the decomposition into partial fractions.

So, $\frac{x}{16 {x}^{4} - 1} = \frac{x}{\left(2 x - 1\right) \left(2 x + 1\right) \left(4 {x}^{2} + 1\right)}$

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

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

so, x=A(2x+1)(4x^2+1)+B(2x-1)(4x^2+1)+(Cx+D)(2x-1)(2x+1))

let $x = - \frac{1}{2}$
$- \frac{1}{2} = B \cdot - 2 \cdot 2$, $\implies$ $B = \frac{1}{8}$

Let $x = \frac{1}{2}$
$\frac{1}{2} = A \cdot 2 \cdot 2$, $\implies$ $A = \frac{1}{8}$

let $x = 0$
$0 = A - B - D$ $\implies$, $D = 0$

Coefficients of ${x}^{3}$
$0 = 8 A + 8 B + 4 C$, $\implies$, $C = - \frac{2}{4} = - \frac{1}{2}$

Therefore,

$\frac{x}{16 {x}^{4} - 1} = \frac{\frac{1}{8}}{2 x - 1} + \frac{\frac{1}{8}}{2 x + 1} + \frac{- \frac{x}{2}}{4 {x}^{2} + 1}$