Precalculus, Chapter 7, 7.4, Section 7.4, Problem 25

[x^2+12x+12]/[x^3-4x]=[x^2+12x+12]/[x(x+2)(x-2)]
[x^2+12x+12]/(x^3-4x)=A/x+B/(x+2)+C/(x-2)

Multiply through by the LCD x^3+4x.
x^2+12x+12=A(x^2-4)+Bx(x-2)+Cx(x+2)
x^2+12x+12=Ax^2-4A+Bx^2-2Bx+Cx^2+2Cx
x^2+12x+12=(A+B+C)x^2+(-2B+2C)x+(-4A)

Equate coefficient of like terms. Then solve for A, B, and C.
1=A+B+C
12=-2B+2C
12=-4A

A=-3

1=A+B+C
1=-3+B+C
4=B+C

Solve for B and C using the elimination method.
4=B+C
12=-2B+2C
Multiply the first equation by 2. Then solve using the elimination method.
8=2B+2C
12=-2B+2C
_________________
20=4C
C=5

1=A+B+C
1=-3+B+5
B=-1

A=-3, B=-1, C=5

(x^2+12x+12)/(x^3-4x)=-3/x-1/(x+2)+5/(x-2)