Your partial fraction will have the form:
\(\displaystyle \frac{Ax+B}{x^{2}+4}+\frac{C}{x-2}=x^{2}+6x\)
Cross multiply the left side:
\(\displaystyle (Ax+B)(x-2)+C(x^{2}+4)=x^{2}+6x\).........[1]
Sub in x=2 because that is what makes x-2=0.
The (Ax+B)(x-2) becomes 0 and we are left with:
\(\displaystyle C(2^{2}+4)=2^{2}+6(2)\)
\(\displaystyle \boxed{C=2}\)
Sub this C value back into [1], set equal to 0, factor a little, and it becomes:
\(\displaystyle (A+1)x^{2}+(-2A+B-6)x-2B+8\)
\(\displaystyle A+1=0, \;\ \boxed{A=-1}\)
\(\displaystyle -2(-1)+B-6=0, \;\ \boxed{B=4}\)
The PFD is:
\(\displaystyle \frac{-x+4}{x^{2}+4}+\frac{2}{x-2}\)
