How do you express $1/((x²)(x² + 3x + 2))$ in partial fractions?

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Answer 1 · 2016-11-15T12:33:37

$1/(x^2(x^2+3x+2))=-3/(4x)+1/(2x^2)+1/(x+1)-1/(4(x+2))$

As $1/(x^2(x^2+3x+2))=1/(x^2(x+2)(x+1))$ , let the partial fractions be given by

$1/(x^2(x+2)(x+1))hArrA/x+B/x^2+C/(x+1)+D/(x+2)$

or $1/(x^2(x+2)(x+1))hArr(Ax(x+1)(x+2)+B(x+1)(x+2)+Cx^2(x+2)+Dx^2(x+1))/(x^2(x+2)(x+1))$

Now, if $x=-1$ , we have $C=1$ and if $x=-2$ , we have $-4D=1$ or $D=-1/4$ and if $x=0$ , we have $2B=1$ or $B=1/2$ .

Now comparing the terms of $x^3$ , we get $A+C+D=0$ and hence
$A=-C-D=-1-(-1/4)=-3/4$

Hence $1/(x^2(x^2+3x+2))=-3/(4x)+1/(2x^2)+1/(x+1)-1/(4(x+2))$

Answer 2 · 2016-11-15T12:51:14

The answer is $=(1/2)/x^2+(-3/4)/x+(-1/4)/(x+2)+(1)/(x+1)$

The denominator $=x^2(x^2+3x+2)=x^2(x+2)(x+1)$

The decomposition in partial fractions is

$1/(x^2(x+2)(x+1))=A/x^2+B/x+C/(x+2)+D/(x+1)$

$=(A(x+1)(x+2)+Bx(x+1)(x+2)+Cx^2(x+1)+Dx^2(x+2))/((x^2)(x+1)(x+2))$

$1=(A(x+1)(x+2)+Bx(x+1)(x+2)+Cx^2(x+1)+Dx^2(x+2))$

Let x=0, $=>$ $1=2A$ ; $A=1/2$

Let $x=-1$ , $=>$ $1=D$

Let $x=-2$ , $=>$ $1=-4C$ , $C=-1/4$

Coefficient of $x^3$ , $=>$ , $0=B+C+D$

$B=-C-D=1/4-1=-3/4$

$1/(x^2(x+2)(x+1))=(1/2)/x^2+(-3/4)/x+(-1/4)/(x+2)+(1)/(x+1)$

How do you express 1/((x²)(x² + 3x + 2)) 1/((x²)(x² + 3x + 2)) in partial fractions?