7. The range of the function $f(x)=x(\sqrt{1+x}+\sqrt{1-x})$ is $\qquad$
7. $\left[-\frac{8 \sqrt{3}}{9}, \frac{8 \sqrt{3}}{9}\right]$.

Obviously, the domain of this function is $[-1,1]$.
$$
\begin{array}{l}
\text { Also, } f(-x)=-x(\sqrt{1-x}+\sqrt{1+x}) \\
=-x(\sqrt{1+x}+\sqrt{1-x})=-f(x),
\end{array}
$$

Thus, the function $f(x)$ is an odd function.
Let $\sqrt{1-x^{2}}=t$. Then
$$
\begin{array}{l}
0 \leqslant t \leqslant 1 \Rightarrow 1-t \geqslant 0 . \\
\text { Hence } f^{2}(x)=x^{2}\left(2+2 \sqrt{1-x^{2}}\right) \\
=2\left(1-t^{2}\right)(1+t) \\
=2(1+t)(1-t)(1+t) \\
=(1+t)(1+t)(2-2 t) \\
\leqslant\left[\frac{(1+t)+(1+t)+(2-2 t)}{3}\right]^{3}=\frac{64}{27},
\end{array}
$$

The equality holds if and only if $3 t=1, x= \pm \frac{2 \sqrt{2}}{3}$.

And $f(0)=0, y=f(x)(0 \leqslant x \leqslant 1)$ has values greater than or equal to zero, so the range of this function is $\left[0, \frac{8 \sqrt{3}}{9}\right]$.
By the property of odd functions, the range of the original function is
$$
\left[-\frac{8 \sqrt{3}}{9}, \frac{8 \sqrt{3}}{9}\right] .
$$