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

1 Answer
Mar 14, 2018

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

Explanation:

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