Example 2 Try to find the unit digit of the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$.

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Notice,
$$
(\sqrt{2}+\sqrt{3})^{2012}=(5+2 \sqrt{6})^{1000} \text {. }
$$

Consider the integer-coefficient quadratic equation $x^{2}-10 x+1=0$ with $5+2 \sqrt{6}$ as one of its roots, the other root being $5-2 \sqrt{6}$.
$$
\begin{array}{l}
\text { Let } a=5+2 \sqrt{6}, b=5-2 \sqrt{6}, \\
u_{n}=a^{n}+b^{n}(n=1,2, \cdots) .
\end{array}
$$

It is easy to see that $a+b=10, a b=1$. 
Since $a^{2}=10 a-1, b^{2}=10 b-1$, for all $n \in \mathbf{N}_{+}$, we have
$$
a^{n+2}=10 a^{n+1}-a^{n}, b^{n+2}=10 b^{n+1}-b^{n} \text {. }
$$

Adding the above two equations gives $u_{n+2}=10 u_{n+1}-u_{n}$, where,
$$
u_{1}=a+b=10, u_{2}=a^{2}+b^{2}=98 \text {. }
$$

Therefore, $u_{n}$ is a positive integer, and
$$
u_{n+4} \equiv-u_{n+2} \equiv u_{n}(\bmod 10)(n=1,2, \cdots) \text {. }
$$

Since $1006=4 \times 251+2$, we get
$$
u_{1006} \equiv u_{2} \equiv 8(\bmod 10) \text {. }
$$

Given $0<b<1$, we know
$$
\left[(\sqrt{2}+\sqrt{3})^{2012}\right]=u_{1000}-1=7(\bmod 10),
$$

which means the unit digit of the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$ is 7.