解析

如图所示：增加一条直线$y = -x + 2$，则$y = -x + 2$将积分区域$D$分为两部分：$D_1$ 和$D - D_1$，其中$D - D_1$关于$y$轴对称，于是

\[

\begin{aligned}

I &= \iint_{D}\frac{x^{2}-2xy + y^{2}}{x^{2}+y^{2}}dxdy \\

&= \iint_{D}\left[1 - \frac{2xy}{x^{2}+y^{2}}\right]dxdy \\

&= \iint_{D}dxdy - \iint_{D}\frac{2xy}{x^{2}+y^{2}}dxdy \\

&= \iint_{D}dxdy - \iint_{D - D_1}\frac{2xy}{x^{2}+y^{2}}dxdy - \iint_{D_1}\frac{2xy}{x^{2}+y^{2}}dxdy \\

&= \pi + 2 - \int_{0}^{\frac{\pi}{2}}d\theta\int_{\frac{2}{\cos\theta+\sin\theta}}^{2}\frac{2\rho^{2}\cos\theta\sin\theta}{\rho^{2}}\rho d\rho \\

&= \pi + 2 - 4\int_{0}^{\frac{\pi}{2}}\cos\theta\sin\theta d\theta + 4\int_{0}^{\frac{\pi}{2}}\frac{\cos\theta\sin\theta}{(\cos\theta+\sin\theta)^{2}}d\theta

\end{aligned}

\]

\[

\begin{aligned}

&= \pi + 4\int_{0}^{\frac{\pi}{2}}\frac{\cos\theta\sin\theta}{1 + 2\cos\theta\sin\theta}d\theta = \pi + 4\int_{0}^{\frac{\pi}{2}}\frac{\tan\theta}{\tan^{2}\theta + 2\tan\theta + 1}d\theta \\

&= \pi + 4\int_{0}^{+\infty}\frac{t}{(t + 1)^{2}}\cdot\frac{1}{1 + t^{2}}dt = \pi + 4\cdot\frac{1}{2}\int_{0}^{+\infty}\left[\frac{1}{1 + t^{2}} - \frac{1}{(t + 1)^{2}}\right]dt \\

&= \pi + 2\arctan t\big|_{0}^{+\infty} + 2\frac{1}{1 + t}\big|_{0}^{+\infty} = \pi + \pi - 2 = 2\pi - 2.

\end{aligned}

\]