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

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Answer 1 · 2016-05-08T06:14:05

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

$(2(1-x-x^2))/(1-x^2)$

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

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

As $(1-x^2)=(1-x)(1+x)$ , let

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

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

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

Hence $A+B=0$ and $A-B=-2$ . Adding them we get

$2A=-2$ or $A=-1$ and $B=1$

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

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