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)#