Question

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

Answer 1

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

Explanation:

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

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

Explanation:

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)`