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

1 Answer
Narad T.
Jan 17, 2017

The answer is =(2/9)/(x+2)+(7/9)/(x-1)+(2/3)/(x-1)^2=29x+2+79x−1+23(x−1)2

Explanation:

Let's do the decomposition into partial fractions

(x^2+x)/((x+2)(x-1)^2)=A/(x+2)+B/(x-1)+C/(x-1)^2x2+x(x+2)(x−1)2=Ax+2+Bx−1+C(x−1)2

=(A(x-1)^2+B(x+2)(x-1)+C(x+2))/((x+2)(x-1)^2)=A(x−1)2+B(x+2)(x−1)+C(x+2)(x+2)(x−1)2

We equalise the denominators

x^2+x=A(x-1)^2+B(x+2)(x-1)+C(x+2)x2+x=A(x−1)2+B(x+2)(x−1)+C(x+2)

Let x=-2x=−2, =>⇒,2=9A2=9A, =>⇒, A=2/9A=29

Let x=1x=1, =>⇒, 2=3C2=3C, =>⇒, C=2/3C=23

Coefficients of x^2x2

1=A+B1=A+B, =>⇒, B=1-A=1-2/9=7/9B=1−A=1−29=79

So,

(x^2+x)/((x+2)(x-1)^2)=(2/9)/(x+2)+(7/9)/(x-1)+(2/3)/(x-1)^2x2+x(x+2)(x−1)2=29x+2+79x−1+23(x−1)2

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