How find this integral $I=\int_{0}^{1}\sqrt{1-W^2(x)}dx$ How find this nice integral

$$I=\int_{0}^{1}\sqrt{1-W^2(x)}dx$$

where, $W(x)$ is  Lambert W function
My try: let $$\sqrt{1-W^2(x)}=u\Longrightarrow W(x)=\sqrt{1-u^2}$$
and since $x=W(x)e^{W(x)}$
so

$$x=W^{-1}(\sqrt{1-u^2})=\sqrt{1-u^2}e^{\sqrt{1-u^2}}?$$
  so
  $$dx=-ue^{1-u^2}\dfrac{1+\sqrt{1-u^2}}{\sqrt{1-u^2}}du$$
  so
  $$I=\int_{1}^{a}-u^2e^{1-u^2}\dfrac{1+\sqrt{1-u^2}}{\sqrt{1-u^2}}du$$

where $a$ such

$$\sqrt{1-a^2}e^{\sqrt{1-a^2}}=1(a>0)$$

then I can't,Thank you very much .

I would make the substitution $u=W(x)$, so $x=ue^u$ and $dx=(u+1)e^u du$, yielding
$$
I=\int_{0}^{\Omega}(u+1)\sqrt{1-u^2}\cdot e^udu;
$$
here $\Omega=W(1)\approx0.573$.  Now expand the square root in powers of $u$:
$$
I=\int_{0}^{\Omega}(u+1)\left(\sum_{k=0}^{\infty}c_ku^{2k}\right)e^u du=\sum_{k=0}^{\infty}c_k\int_{0}^{\Omega}(u^{2k}+u^{2k+1})e^u du =\sum_{k=0}^{\infty}c_k u^{2k+1}\left(\Gamma(2k+1,-u)-\Gamma(2k+2,-u)\right)\big\vert_{0}^{\Omega}=\sum_{k=0}^{\infty}c_k\left(\Omega^{2k+1}\Gamma(2k+1,-\Omega)-\Omega^{2k+2}\Gamma(2k+2,-\Omega)\right)=\sum_{k=1}^{\infty}d_k\Omega^{k},
$$
where
$$
c_k=\frac{1}{k!}\frac{d^k}{dx^k}\sqrt{1-x}=(-1)^k\left(\frac{3}{2}-k\right)_{k}
$$
and
$$
d_k=(-1)^{k-1} c_{\lfloor{(k-1)/2}\rfloor}\Gamma(k,-\Omega).
$$
Putting these together,
$$
I=\sum_{k=1}^{\infty}(-1)^{k-1+\lfloor{(k-1)/2}\rfloor}\left(\frac{3}{2}-\left\lfloor{\frac{k-1}{2}}\right\rfloor\right)_{\lfloor{(k-1)/2}\rfloor}\Gamma(k,-\Omega)\Omega^{k}.
$$
