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

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Answer 1 · 2017-11-15T08:20:39

The answer is #=-1/(x-1)^2+1/(x-1)-((x-1))/(x^2+1)#

Perform the decomposition into partial fractions

#(x^2-2x-1)/((x-1)^2(x^2+1))=A/(x-1)^2+B/(x-1)+(Cx+D)/(x^2+1)#

#=(A(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)^2)/((x-1)^2(x^2+1))#

The denominators are the same, compare the numerators

#(x^2-2x-1)=A(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x-1)^2#

Let #x=1#

#-2=2A#, #A=-1#

Coefficients of #x^2#

#1=A-B-2C+D#

Coefficients of #x#

#-2=B+C-2D#

and

#-1=A-B+D#, #=>#, #-B+D=0#, #B=D#

#1=-1-B-2C+B#, #=>#, #2C=-2#, #=>#, #C=-1#

#-2=B-1-2B#, #=>#, #B=1#

#=>#, #D=1#

Finally,

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