Answer
$\frac{1}{x-2}+\frac{1}{(x-2)^2}-\frac{1}{x+2}-\frac{2}{(x+2)^2}$
Work Step by Step
Step 1. Factor the denominator as: $(x^2-4)^2=(x+2)^2(x-2)^2$
Step 2. Assume the end results as: $\frac{A}{x+2}+\frac{B}{(x+2)^2}+\frac{C}{x-2}+\frac{D}{(x-2)^2}$
Step 3. Combine the above functions as:
$\frac{A(x+2)(x-2)^2+B(x-2)^2+C(x-2)(x+2)^2+D(x+2)^2 }{(x+2)^2(x-2)^2}=\frac{(A+C)x^3+(-2A+B+2C+D)^2+(-4A-4B-4C+4D)x+(8A+4B-8C+4D) }{(x+2)^2(x-2)^2}$
Step 4. Compare the above result with the original expression to set up the following system of equations:
\begin{cases} A+C=0 \\ -2A+B+2C+D=3\\-4A-4B-4C+4D=12\\8A+4B-8C+4D=-20 \end{cases}
Step 5. Use substitution A=-C to get:
\begin{cases} B+4C+D=3\\-4B+4D=12\\4B-16C+4D=-20 \end{cases}
Step 6. The second equation gives D=B+3, use it for substitution to get:
\begin{cases} 2B+4C=0\\2B-4C=-8 \end{cases}
Step 7. Solve the above equations and other parameters to get $A=-1,B=-2,C=1,D=1$
Step 8. Write the final results as:
$-\frac{1}{x+2}-\frac{2}{(x+2)^2}+\frac{1}{x-2}+\frac{1}{(x-2)^2}$ or
$\frac{1}{x-2}+\frac{1}{(x-2)^2}-\frac{1}{x+2}-\frac{2}{(x+2)^2}$
