5. The area enclosed by the curve $x^{2}+y^{2}-|x|-|y|=0$ is $\qquad$ .
5. When $x \geqslant 0, y \geqslant 0$, the given equation is $x^{2}+y^{2}-x-y=0$, which can be rewritten as $\left(x-\frac{1}{2}\right)^{2}+\left(y-\frac{1}{2}\right)^{2} = \frac{1}{2}$. The corresponding curve in the first quadrant (including endpoints) is a semicircle with its center at $\left(\frac{1}{2}, \frac{1}{2}\right)$ and radius $\frac{\sqrt{2}}{2}$.

Similarly, when $x \leqslant 0, y \geqslant 0$; $x \leqslant 0, y \leqslant 0$; and $x \geqslant 0, y \leqslant 0$, the corresponding curves are all semicircles (as shown in the figure). The area enclosed by these curves is $4\left[\frac{1}{2} \cdot 1 \cdot 1\right.$
$$
\begin{array}{l}
\left.+\frac{1}{2} \cdot \pi \cdot\left(\frac{\sqrt{2}}{2}\right)^{2}\right] \\
=2+\pi .
\end{array}
$$