= Solution
Use the suggested substitution
$$
x=\frac{t}{\sqrt{2-t^2}}.
$$
It maps $[0,1]$ to itself and satisfies
$$
dx=\frac{2\,dt}{(2-t^2)^{3/2}},
\qquad
1-x^4=\frac{4(1-t^2)}{(2-t^2)^2}.
$$
Hence
$$
\int_0^1\frac{dx}{\sqrt{1-x^4}}
=\frac1{\sqrt2}\int_0^1
\frac{dt}{\sqrt{(1-t^2)(1-t^2/2)}}
=\frac1{\sqrt2}K(1/\sqrt2),
$$
where $K$ is the <complete elliptic integral of the first kind>. Part (a) and the <Gamma function recurrence> $\Gamma(5/4)=\Gamma(1/4)/4$ now give
$$
\boxed{K(1/\sqrt2)
=\frac{\Gamma(1/4)^2}{4\sqrt\pi}
=\frac{4\Gamma(5/4)^2}{\sqrt\pi}}.
$$
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