$\displaystyle I=\int_{0}^1 \sqrt{1-x^4}dx$

Let $\displaystyle t=x^4$ so $\displaystyle dt=4x^3dx$ this is $\displaystyle x=t^{1/4}$ and $\displaystyle dx=\frac{dt}{4x^3}=\frac{dt}{4t^{3/4}}$

$\displaystyle \int_{0}^1(1-t)^{1/2} t^{-3/4}dt=\int_{0}^1(1-t)^{1/2} t^{-3/4}dt$

The definition of the beta function reads

$\displaystyle B(x,y) = \int_0^1t^{x-1}(1-t)^{y-1}\,dt$

so that $\displaystyle y-1=\frac{1}{2}$ and $\displaystyle x-1=\frac{3}{4}$ hence $\displaystyle y=\frac{3}{2}, x=\frac{7}{4}$

so

$\displaystyle \int_{0}^1(1-t)^{1/2} t^{-3/4}dt=B(\frac{7}{4},\frac{3}{2})$

Using that

$\displaystyle B(x,y)=\dfrac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}$

my answer is

$\displaystyle 4I=\frac{\Gamma(\frac{7}{4})\Gamma( \frac{3}{2})}{ \Gamma (\frac{13}{4})}$

which is not yours either