How do I find the partial fraction decomposition of (1)/(x^3+2x^2+x1x3+2x2+x ?

1 Answer
Cem Sentin
Mar 14, 2018

1/x-1/(x+1)-1/(x+1)^21x−1x+1−1(x+1)2

Explanation:

1/(x^3+2x^2+x)=1/[x*(x^2+2x+1)]=1/[x*(x+1)^2]1x3+2x2+x=1x⋅(x2+2x+1)=1x⋅(x+1)2

Hence,

1/(x^3+2x^2+x)=A/x+B/(x+1)+C/(x+1)^21x3+2x2+x=Ax+Bx+1+C(x+1)2

After expanding denominator,

A*(x+1)^2+B*(x^2+x)+Cx=1A⋅(x+1)2+B⋅(x2+x)+Cx=1

Set x=-1x=−1, -C=1−C=1, so C=-1C=−1

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

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

Thus,

1/(x^3+2x^2+x)=1/x-1/(x+1)-1/(x+1)^21x3+2x2+x=1x−1x+1−1(x+1)2

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