How do you write the partial fraction decomposition of the rational expression $(x^2 + 2) /( x^3 + 3x^2 + 3x + 1)$ ?

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Answer 1 · 2016-11-09T07:53:36

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

The denominator is $(x+1)^3$
$:.(x^2+2)/(x+1)^3=A/(x+1)^3+B/(x+1)^2+C/(x+1)$
$=(A+B(x+1)+C(x+1)^2)/(x+1)^3$

$:.x^2+2=A+B(x+1)+C(x+1)^2$

let $x=-1$ $=>$ $3=A$

If $x=0$ , $2=A+B+C$
coefficients of $x^2$ $=>$ $1=C$
coefficients of $x$ $=>$ $0=B+2C$ $=>$ $B=-2$

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

How do you write the partial fraction decomposition of the rational expression (x^2 + 2) /( x^3 + 3x^2 + 3x + 1)(x^2 + 2) /( x^3 + 3x^2 + 3x + 1)?