8x/((x-2)²(x+2))≡A/(x-2)+B/(x-2)²+C/(x+2)
So 8x≡A(x²-4)+B(x+2)+C(x²-4x+4)=(A+C)x²+(B-4C)x+(-4A+2B+4C).
So A+C=0, C=-A; B-4C=B+4A=8, B=8-4A; -4A+16-8A-4A=0, -16A+16=0, A=1.
Therefore C=-1, B=8-4=4.
So the partial fractions are: 1/(x-2)+4/(x-2)²-1/(x+2).