Answer
$\frac{1}{x-2}-\frac{1}{x+2}-\frac{4}{(x+2)^2}$
Work Step by Step
1.Write a general form $\frac{16}{(x-2)(x+2)^2}=\frac{A}{x-2}+\frac{B}{x+2}+\frac{C}{(x+2)^2}$
2. Combine the right part to get $\frac{Ax^2+4Ax+4A+Bx^2-4B+Cx-2C}{(x-2)(x+2)^2}=\frac{(A+B)x^2+(4A+C)x+4A-4B-2C}{(x-2)(x+2)^2}$
3. Compare with the starting expression, we have
$\begin{cases} A+B=0 \\ 4A+C=0 \\ 4A-4B-2C=16 \end{cases}$
4. Use the relations $B=-A, C=-4A$ in the third equation to get
$4A+4A+8A=16$, thus $A=1, B=-1, C=-4$,
5. Thus $\frac{16}{(x-2)(x+2)^2}=\frac{1}{x-2}-\frac{1}{x+2}-\frac{4}{(x+2)^2}$
