Example 11 Express $\frac{x}{(x-2)\left(x^{2}+2 x+2\right)}$ as partial fractions.
Solution: Since the discriminant of $x^{2}+2 x+2$ in the denominator is less than zero, it cannot be factored into a product of linear factors. We call it a quadratic prime factor. Let
$$
\begin{array}{l}
N=\frac{x}{(x-2)\left(x^{2}+2 x+2\right)} \\
=\frac{A}{x-2}+\frac{B x+C}{x^{2}+2 x+2} .
\end{array}
$$

Then $x=A\left(x^{2}+2 x+2\right)+(B x+C)(x-2)$.
Let $x=2$, substituting we get $A=\frac{1}{5}$.
Comparing the coefficients of $x^{2}$ on both sides of the identity, we get $0=\frac{1}{5}+$ $B$, then $B=-\frac{1}{5}$; comparing the constant terms, we get $0=2 A-2 C$, then $C=A=\frac{1}{5}$.
Therefore, $N=\frac{1}{5(x-2)}+\frac{-x+1}{5\left(x^{2}+2 x+2\right)}$.