Example 4 Try to find the unit digit of the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$.
[2]
Notice that,
$$
(\sqrt{2}+\sqrt{3})^{2012}=(5+2 \sqrt{6})^{1006} \text {. }
$$

Clearly, $0<(5-2 \sqrt{6})^{1006}<1$,
$$
\begin{array}{l}
(5+2 \sqrt{6})^{1006}+(5-2 \sqrt{6})^{1006} \\
=2\left(C_{1006}^{0} 5^{1006}+C_{1006}^{2} 5^{1004} \times 24+\cdots+\right. \\
\left.\quad C_{1006}^{1006} 24^{503}\right) \in \mathbf{Z}_{+} .
\end{array}
$$

Therefore, the integer part of $(\sqrt{2}+\sqrt{3})^{2012}$ is
$$
2\left(\mathrm{C}_{1006}^{0} 5^{1006}+\mathrm{C}_{1006}^{2} 5^{1004} \times 24+\cdots+\mathrm{C}_{1006}^{100} 24^{503}\right)-1 \text {. }
$$

And $2\left(\mathrm{C}_{1006}^{0} 5^{1006}+\mathrm{C}_{1006}^{2} 5^{1004} \times 24+\cdots+\right.$
$$
\begin{array}{l}
\left.C_{1006}^{1006} 24^{503}\right)-1 \\
\equiv 2 \times 24^{503}-1 \equiv 2 \times(25-1)^{503}-1 \\
\equiv 2 \times(-1)^{503}-1 \equiv 7(\bmod 10),
\end{array}
$$

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