How do you express $(x^3 +x^2+2x+1) / [(x^2+1) (x^2+2)]$ in partial fractions?

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Answer 1 · 2017-05-26T15:29:44

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

Let's perform the decomposition into partial fractions

$(x^3+x^2+2x+1)/((x^2+1)(x^2+2))=(Ax+B)/(x^2+1)+(Cx+D)/(x^2+2)$

$=((Ax+B)(x^2+1)+(Cx+D)(x^2+1))/((x^2+1)(x^2+2))$

The denominators are the same, we compare the numerators

$x^3+x^2+2x+1=(Ax+B)(x^2+2)+(Cx+D)(x^2+1)$

Coeficients of $x^3$

$1=A+C$

Let $x=0$ , $=>$ , $1=2B+D$

Coefficients of $x^2$

$1=B+D$

Coefficients of $x$

$2=2A+C$

Solving for $A$ , $B$ , $C$ and $D$ from the 4 equations

$2=2A+1-A$ , $=>$ , $A=1$

$C=1-A=1-1=0$

$1=2B+1-B$ , $=>$ , $B=0$

$1=B+D$ , $=>$ , $D=1$

Therefore,

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

How do you express (x^3 +x^2+2x+1) / [(x^2+1) (x^2+2)] (x^3 +x^2+2x+1) / [(x^2+1) (x^2+2)] in partial fractions?