How do you express $(x^4)/((x²+1)(x²-1))$ in partial fractions?

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Answer 1 · 2016-11-30T18:19:21

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

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

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

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

$=(Ax+B)/(x^2+1)+C/(x+1)+D/(x-1)$

$=((Ax+B)(x^2-1)+C(x-1)(x^2+1)+D(x+1)(x^2+1))/((x^2+1)(x^2-1))$

$1=(Ax+B)(x^2-1)+C(x-1)(x^2+1)+D(x+1)(x^2+1))$

Let $x=1$ , $=>$ , $1=4D$ , $=>$ , $D=1/4$

Let $x=-1$ , $=>$ , $1=-4C$ , $=>$ , $C=-1/4$

No coefficients, $1=-B-C+D$ , $=>$ , $B=1/2-1=-1/2$

Coefficients of $x^3$ , $=>$ , $0=A+C+D$ , $=>$ . $A=0$

Therefore,

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

How do you express (x^4)/((x²+1)(x²-1))(x^4)/((x²+1)(x²-1)) in partial fractions?