Example 10 Express $\frac{2 x^{2}+1}{x^{2}(x-1)^{2}}$ as partial fractions.
Solution: Let $N=\frac{2 x^{2}+1}{x^{2}(x-1)^{2}}$
$$
\equiv \frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x-1}+\frac{D}{(x-1)^{2}} \text {. }
$$

Then $2 x^{2}+1$
$$
\begin{aligned}
\equiv & A x(x-1)^{2}+B(x-1)^{2}+ \\
& C x^{2}(x-1)+D x^{2} .
\end{aligned}
$$

Let $x=0$, we get $B=1$;
Let $x=1$, we get $D=3$.
Substitute and rearrange to get
$$
-2 x^{2}+2 x \equiv A x(x-1)^{2}+C x^{2}(x-1) \text {. }
$$

Divide both sides of the identity by $x(x-1)$ to get
$$
-2 \equiv A(x-1)+C x \text {. }
$$

Let $x=0$, we get $A=2$;
Let $x=1$, we get $C=-2$.
Therefore, $N=\frac{2}{x}+\frac{1}{x^{2}}+\frac{-2}{x-1}+\frac{3}{(x-1)^{2}}$.