Prove that : $\int_0^{1}\frac{1-\sqrt{1-x^{4}}}{x^{2}\sqrt{1-x^4}}dx=1-\frac{\sqrt{2}\pi^{\frac{3}{2}}}{\Gamma(\frac{1}{4})^2}$ Prove that :
$$\int_0^{1}\frac{1-\sqrt{1-x^{4}}}{x^{2}\sqrt{1-x^4}}dx=1-\frac{\sqrt{2}\pi^{\frac{3}{2}}}{\Gamma(\frac{1}{4})^2}.$$
I know how to use the definition of the Beta and Gamma functions so my problem here is when I divide the integral I find:
$$\int_0^{1}\frac{1}{x^2}dx=\infty!!$$ 
(divergent)
I tried using $y=\frac{1}{x}$ but didn't get the answer.
I also tried $y=x^{4}$ and got the same problem (a divergent integral).

I begin by making the following observation:
$$\frac{d}{dx} \left (\frac{1 - \sqrt{1 - x^4}}{x} \right )= \frac{x^4 + 1 - \sqrt{1 - x^4}}{x^2 \sqrt{1 - x^4}}.$$
So, for your integral, one can write
\begin{align}
\int_0^1 \frac{1 - \sqrt{1 - x^4}}{x^2 \sqrt{1 - x^4}} \, dx &= \int_0^1 \frac{(x^4 + 1 - \sqrt{1 - x^4}) - x^4}{x^2 \sqrt{1 - x^4}} \, dx\\
&= \int_0^1 \left (\frac{1 - \sqrt{1 - x^4}}{x} \right )' \, dx -\underbrace{\int_0^1 \frac{x^2}{\sqrt{1 - x^4}} \, dx}_{x \, \mapsto \, \sqrt[4]{x}}\\
&= 1 - \frac{1}{4} \int_0^1 x^{-\frac{1}{4}} (1 - x)^{-\frac{1}{2}} \, dx\\[1ex]
&= 1 - \frac{1}{4} \operatorname{B} \left (\frac{3}{4}, \frac{1}{2} \right )\\[1ex]
&= 1 - \frac{1}{4} \frac{\Gamma (\frac{3}{4}) \Gamma (\frac{1}{2})}{\Gamma (\frac{5}{4})}\\
&= 1 - \sqrt{\pi} \frac{\Gamma (\frac{3}{4})}{\Gamma (\frac{1}{4})}\\
&= 1 - \frac{\pi \sqrt{2 \pi}}{\Gamma^2 (\frac{1}{4})},
\end{align}
as required. Here $\Gamma (\frac{3}{4}) \Gamma (\frac{1}{4}) = \pi \sqrt{2}$, has been used. 
