3. The range of the function $f(x)=\sqrt{2 x-x^{2}}+x$ is
3. $[0, \sqrt{2}+1]$.

It is easy to know, $f(x)=\sqrt{1-(x-1)^{2}}+x$.
Let $x-1=\sin \alpha\left(-\frac{\pi}{2} \leqslant \alpha \leqslant \frac{\pi}{2}\right)$. Then
$$
\begin{array}{l}
f(x)=\cos \alpha+(1+\sin \alpha) \\
=\sqrt{2} \sin \left(\alpha+\frac{\pi}{4}\right)+1
\end{array}
$$

From $-\frac{\pi}{2} \leqslant \alpha \leqslant \frac{\pi}{2}$
$$
\begin{array}{l}
\Rightarrow-\frac{\pi}{4} \leqslant \alpha+\frac{\pi}{4} \leqslant \frac{3 \pi}{4} \\
\Rightarrow-\frac{\sqrt{2}}{2} \leqslant \sin \left(\alpha+\frac{\pi}{4}\right) \leqslant 1 .
\end{array}
$$

Therefore, the range of $f(x)$ is $[0, \sqrt{2}+1]$.