Answer
$\frac{1}{x+2}+\frac{1}{x-2}-\frac{2}{x}$
Work Step by Step
Step 1. Factorize the denominator as: $x(x^2-4)=x(x+2)(x-2)$
Step 2. Assume the partial fraction decomposition is:
$\frac{A}{x}+\frac{B}{x+2}+\frac{C}{x-2}$
Step 3. Combine the rational functions above:
$\frac{A(x^2-4)+Bx(x-2)+Cx(x+2)}{x(x+2)(x-2)}=\frac{(A+B+C)x^2+(-2B+2C)x+(-4A)}{x(x+2)(x-2)}$
Step 4. Compare the above function with the original to setup the following system of equations:
$\begin{cases} A+B+C=0\\-2B+2C=0\\-4A=8 \end{cases}$
Step 5. Solve the above equations: use $A=-2, B=C$ to plug-in the first equation, $-2+2B=0$ or $B=1$, so $C=1$
Step 6. The final result is:
$\frac{-2}{x}+\frac{1}{x+2}+\frac{1}{x-2}$ or $\frac{1}{x+2}+\frac{1}{x-2}-\frac{2}{x}$
