= Solution
With $u=x^4$, the <change of variables formula> gives
$$
\int_0^1\frac{dx}{\sqrt{1-x^4}}
=\frac14\int_0^1u^{-3/4}(1-u)^{-1/2}\,du
=\frac14B\left(\frac14,\frac12\right).
$$
The <beta--gamma identity>, $\Gamma(1/2)=\sqrt\pi$, and the <Gamma reflection formula> give
$$
\frac14B\left(\frac14,\frac12\right)
=\frac{\Gamma(1/4)\sqrt\pi}{4\Gamma(3/4)},
\qquad
\Gamma(1/4)\Gamma(3/4)=\pi\sqrt2.
$$
Therefore
$$
\boxed{\int_0^1\frac{dx}{\sqrt{1-x^4}}
=\frac{\Gamma(1/4)^2}{\sqrt{32\pi}}}.
$$
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