Answer
$x+1+\frac{1}{(x-2)^2}+\frac{2x+8}{(x-2)^2(x^2+2)}$
Work Step by Step
Step 1. Use long division (see page 269), first divide the numerator by $(x^2+2)$, the original function becomes
$\frac{1}{(x-2)^2}(x^3-3x^2+x+2+\frac{2x+8}{x^2+2})$
Step 2. Use synthetic divisions (see page 270) to divide the polynomial part by the factor $(x-2)$ twice, the above expression becomes
$x+1+\frac{1}{(x-2)^2}+\frac{2x+8}{(x-2)^2(x^2+2)}$
Step 3. We examine the above results and found that no further steps will be necessary.
