How do you express $(x^2+x+1)/(1-x^2)$ in partial fractions?

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Answer 1 · 2016-05-08T07:14:48

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

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

$(x^2+x+1)/(1-x^2)=-1+(x+2)/(1-x^2)$ and let

$(x+2)/(1-x^2)=(x+2)/((1-x)(1+x))hArrA/(1-x)+B/(1+x)$ or

$(x+2)/((1-x)(1+x))hArr(A(1+x)+B(1-x))/((1-x)(1+x))$ or

$(x+2)/((1-x)(1+x))hArr((A-B)x+(A+B))/((1-x)(1+x))$ or

Hence $A-B=1$ and $A+B=2$ or $A=3/2$ and $B=1/2$

Hence $(x^2+x+1)/(1-x^2)=-1+3/(2(1-x))+1/(2(1+x))$

How do you express (x^2+x+1)/(1-x^2) (x^2+x+1)/(1-x^2) in partial fractions?