partial fraction

Resolve [4x(x+4)]/[(x^2－4)(x+2)] into a partial fraction
Sol
[4x(x+4)]/[(x^2－4)(x+2)]=[4x(x+4)]/[(x－2)(x+2)^2]
Set
[4x(x+4)]/[(x^2－4)(x+2)]=a/(x－2)+b/(x+2)+c/(x+2)^2
4x(x+4)=a(x+2)^2+b(x－2)(x+2)+c(x－2)
when x=－2
4*(－2)*2=c(－2－2)
c=4
4x^2+16x=a(x+2)^2+b(x－2)(x+2)+4x－8
4x^2+12x+8=a(x+2)^2+b(x－2)(x+2)
4x+4=a(x+2)+b(x－2)
when x=－2
－8+4=b(－2－2)
b=1
4x+4=a(x+2)+x－2
3x+6=a(x+2)
a=3
So
[4x(x+4)]/[(x^2－4)(x+2)]=3/(x－2)+1/(x+2)+4/(x+2)^2