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

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Answer 1 · 2016-11-20T08:57:38

The answer is $=(1/8)/(2x-1)+(1/8)/(2x+1)+(-x/2)/(4x^2+1)$

Let's factorise the denominator

$16x^4-1=(4x^2-1)(4x^2+1)=(2x-1)(2x+1)(4x^2+1)$

We can start the decomposition into partial fractions.

So, $x/(16x^4-1)=x/((2x-1)(2x+1)(4x^2+1))$

$=A/(2x-1)+B/(2x+1)+(Cx+D)/(4x^2+1)$

$=(A(2x+1)(4x^2+1)+B(2x-1)(4x^2+1)+(Cx+D)(2x-1)(2x+1))/((2x-1)(2x+1)(4x^2+1))$

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

let $x=-1/2$
$-1/2=B*-2*2$ , $=>$ $B=1/8$

Let $x=1/2$
$1/2=A*2*2$ , $=>$ $A=1/8$

let $x=0$
$0=A-B-D$ $=>$ , $D=0$

Coefficients of $x^3$
$0=8A+8B+4C$ , $=>$ , $C=-2/4=-1/2$

Therefore,

$x/(16x^4-1)=(1/8)/(2x-1)+(1/8)/(2x+1)+(-x/2)/(4x^2+1)$

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