Answer
$\displaystyle \frac{2}{15}(8\sqrt{2}-7)\approx 0.57516$
Work Step by Step
Substituting so that so $u^{2}=1-x^{2},\ \quad (u=\sqrt{1-x^{2}})$
$2udu=-2xdx$,
$xdx=-udu,$
the integral becomes simpler
bounds:$\left[\begin{array}{l}
x=0\rightarrow u=1\\
x=1\rightarrow u=0
\end{array}\right]$:
$I=\displaystyle \int_{0}^{1}x\sqrt{2-\sqrt{1-x^{2}}}dx=\int_{1}^{0}\sqrt{2-u}(-udu)$
$\left[\begin{array}{lll}
v=\sqrt{2-u} & v^{2}=2-u & u=2-v^{2}\\
& 2vdv=-du & \\
& &
\end{array}\right] \left[\begin{array}{l}
u=0\rightarrow v=\sqrt{2}\\
u=1\rightarrow v=1
\end{array}\right]$
$I=\displaystyle \int_{1}^{\sqrt{2}}v(2-v^{2})(2vdv)$
$=\displaystyle \int_{1}^{\sqrt{2}}(4v^{2}-2v^{4})dv$
... both are table integrals type 1
$=[\displaystyle \frac{4}{3}v^{3}-\frac{2}{5}v^{5}]_{1}^{\sqrt{2}}$
$=(\displaystyle \frac{4}{3}2^{3/2}-\frac{2}{5}2^{5/2})-(\frac{4}{3}-\frac{2}{5})$
$=(\displaystyle \frac{8}{3}2^{1/2}-\frac{8}{5}2^{1/2})-(\frac{14}{15})$
$=\displaystyle \frac{16}{15}2^{1/2}-\frac{14}{15}$
$=\displaystyle \frac{2}{15}(8\sqrt{2}-7)\approx 0.57516$