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

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Answer 1 · 2017-01-20T10:35:21

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

Let's perform the decomposition into partial fractions

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

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

Equalising the numerators

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

Let #x=0#, #=>#, #1=A#

Coefficients of #x#, #1=B#

Coefficients of #x^2#, #0=2A+D+F#

Coefficients of #x^3#, #0=2B+C+E#

coefficients of #x^4#, #0=A+F#, #=>#, #F=-A=-1#

#D=-2A-F=-2+1=-1#

Coefficients of #x^5#, #0=B+E#, #=>#, #E=-B=-1#

#C=-2B-E=-2+1=-1#

Therefore,

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