Answer
$\frac{2}{x^2+4}-\frac{x}{(x^2+4)^2}$
Work Step by Step
Step 1. The denominator can not be further factorized, assume the end results as:
$\frac{Ax+B}{x^2+4}+\frac{Cx+D}{(x^2+4)^2}$
Step 2. Combine the above functions as:
$\frac{(Ax+B)(x^2+4)+Cx+D}{(x^2+4)^2}=\frac{(A)x^3+(B)x^2+(4A+C)x+(4B+D)}{(x^2+4)^2}$
Step 3. Compare the above result with the original expression to set up the following system of equations:
\begin{cases} A=0 \\ B=2\\4A+C=-1\\4B+D=8 \end{cases}
Step 4. Use substitution to solve the above equations and get $A=0, B=2,C=-1,D=0$
Step 5. Write the final results as:
$\frac{2}{x^2+4}-\frac{x}{(x^2+4)^2}$
