Question

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

Answer 1

`(x^2+x)/((x+2)(x-1)^2)=2/(9(x+2))+7/(9(x-1))+2/(3(x-1)^2)`

Explanation:

As the degree of numerator is less than that of denominator, we can proceed to write partial fractions as

`(x^2+x)/((x+2)(x-1)^2)hArrA/(x+2)+B/(x-1)+C/(x-1)^2`

or `x^2+x=A(x-1)^2+B(x-1)(x+2)+C(x+2)`............(1)

or `x^2+x=A(x^2-2x+1)+B(x^2+x-2)+C(x+2)`

or `x^2+x=x^2(A+B)+x(B-2A+C)+(A-2B+2C)`............(2)

Putting `x=1` and `x=-2` in (1), we get `3C=2` and `9A=2` i.e. `C=2/3` and `A=2/9`.

Comparing terms of `x^2` in (2), `A+B=1` i.e. `B=7/9`

Hence `(x^2+x)/((x+2)(x-1)^2)=2/(9(x+2))+7/(9(x-1))+2/(3(x-1)^2)`