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,

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

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