How do you express $(x^5 + 1)/(x^6 - x^4)$ in partial fractions?

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Answer 1 · 2016-03-08T18:02:33

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

$(x^5+1)/(x^4(x^2-1)) =(x^5+1)/(x^4(x+1)(x-1))$
$(x^5+1)/(x^4(x+1)(x-1))= A/x +B/x^2+C/x^3+D/x^4 +E/(x+1)+F/(x-1)$

$x^5+1=A(x^3)(x^2-1)+B(x^2)(x^2-1)+C(x)(x^2-1)+D(x^2-1)+E(x^4)(x-1)+F(x^4)(x+1)$

$x^5+1=Ax^5-Ax^3+Bx^4-Bx^2+Cx^3-Cx+Dx^2-D+Ex^5-Ex^4+Fx^5+Fx^4$

$1=A+E+F, 0=B-E+F, 0=-A+C,0=-B+D, 0=-C, 1=-D$

$A=0,B=-1,C=0,D=-1,E=0,F=1$

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

How do you express (x^5 + 1)/(x^6 - x^4)(x^5 + 1)/(x^6 - x^4) in partial fractions?