How do you express $(x^3 + 2) / (x^4 -16)$ in partial fractions?

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Answer 1 · 2016-11-20T07:20:49

The answer is $=(3/16)/(x+2)+(5/16)/(x-2)+(x/2-1/4)/(x^2+4)$

Let's factorise the denominator

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

Therefore,

$(x^3+2)/(x^4-16)=(x^3+2)/((x+2)(x-2)(x^2+4))$

The decomposition in partial fractions is

$(x^3+2)/(x^4-16)=A/(x+2)+B/(x-2)+(Cx+D)/(x^2+4)$

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

Therefore,

$(x^3+2)=(A(x-2)(x^2+4)+B(x+2)(x^2+4)+(Cx+D)(x-2)(x+2))$

let $x=2$
$10=32B$ ; $=>$ $B=5/16$

Let $x=-2$
$-6=-32A$ ; $=>$ $A=3/16$

Coefficients of $x^3$
$1=A+B+C$ , $=>$ $C=1-5/16-3/16=8/16=1/2$

Let $x=0$
$2=-8A+8B-4D$ : $=>$ $4D=-2-3/2+5/2=-1$
$D=-1/4$

$(x^3+2)/(x^4-16)=(3/16)/(x+2)+(5/16)/(x-2)+(x/2-1/4)/(x^2+4)$

How do you express (x^3 + 2) / (x^4 -16) (x^3 + 2) / (x^4 -16) in partial fractions?