How do I find the partial fraction decomposition of #(1)/(x^3+2x^2+x# ?

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Answer 1 · 2018-03-14T23:54:46

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

#1/(x^3+2x^2+x)=1/[x*(x^2+2x+1)]=1/[x*(x+1)^2]#

Hence,

#1/(x^3+2x^2+x)=A/x+B/(x+1)+C/(x+1)^2#

After expanding denominator,

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

Set #x=-1#, #-C=1#, so #C=-1#

Set #x=0#, #A=1#

Set #x=1#, #-4A+2B+C=1#, so #B=-1#

Thus,

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