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

1 Answer
May 8, 2016

Answer:

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

Explanation:

#(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))#