How to tackle the integral $\int_{0}^{1} \sqrt{-1+\sqrt{\frac{4}{x}-3}} d x$? $ \text {Let } y=\sqrt{-1+\sqrt{\frac{4}{x}-3}}\textrm{ then ,}$
$ \displaystyle \begin{aligned}I&=16 \int_{0}^{\infty} \frac{y^{2}\left(y^{2}+1\right) d y}{\left(y^{4}+2 y^{2}+4\right)^{2}}\\&=4\left[3 \underbrace{\int_{0}^{\infty} \frac{y^{2}\left(y^{2}+2\right)}{\left(y^{4}+2 y^{2}+4\right)^{2}} d y}_{J}+\underbrace{\int_{0}^{\infty} \frac{y^{2}\left(y^{2}-2\right)}{\left(y^{4}+2 y^{2}+4\right)^{2}}}_{K} d y\right] \end{aligned}\tag*{} $
Now let’s play a little trick on the integral $ J$.
$\displaystyle \begin{aligned}J &=\int_{0}^{\infty} \frac{1+\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y^{2}}+2\right)^{2}} d y \\&=\int_{0}^{\infty} \frac{d\left(y-\frac{2}{y}\right)}{\left[\left(y-\frac{2}{y}\right)^{2}+6\right]^{2}} \\&=\int_{-\infty}^{\infty} \frac{d u}{\left(u^{2}+6\right)^{2}}\\ &\stackrel{u=\sqrt6 \tan \theta}{=}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sqrt{6} \sec ^{2} \theta d \theta}{\left(6 \sec ^{2} \theta\right)^{2}}\\&=\frac{\pi}{12 \sqrt{6}} \end{aligned} \tag*{} $
For the integral $ K$ , we first split the interval into two.
$ \displaystyle \begin{aligned}K &=\int_{0}^{\infty} \frac{1-\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y^{2}}+2\right)^{2}} d y \\&=\int_{0}^{1} \frac{1-\frac{1}{y^{2}}}{\left(y^{2}+\frac{4}{y^{2}}+2\right)^{2}} d y+\int_{1}^{\infty} \frac{1-\frac{2}{y^{2}}}{\left(y^{2}+\frac{4}{y}+2\right)^{2}} d y \\&=\int_{0}^{1} \frac{d\left(y+\frac{2}{y}\right)}{\left[\left(y+\frac{2}{y}\right)^{2}-2\right]^{2}}+\int_{3}^{\infty} \frac{d\left(y+\frac{2}{y}\right)}{\left[\left(y+\frac{2}{y}\right)^{2}-2\right]^{2}} d y \\&=\int_{\infty}^{3} \frac{d u}{\left(u^{2}-2\right)^{2}}+\int_{3}^{\infty} \frac{d v}{\left(v^{2}-2\right)^{2}} \\&=0 \end{aligned} \tag*{} $
Now we can conclude that
$\displaystyle \boxed{I=4\left(3 \cdot \frac{\pi}{12 \sqrt{6}}\right)=\frac{\pi}{\sqrt{6}}}\tag*{} $
My Question
Is there any other substitution or method to tackle the integral?

Taking the inverse function $x=x(y)$ and integrating between $0$ and $\infty$ gives that:
$\int_0^\infty \frac{4}{y^{4}+2y^{2}+4}dy$,
Decomposing the rational function into the sum of several simple fractions gives:
$\frac{4}{(y^{2}+\sqrt{2}y+2)( y^{2}-\sqrt{2}y+2)}=$,
$\frac{A y+B}{ y^{2}+\sqrt{2}y+2}+\frac{C y+D}{ y^{2}-\sqrt{2}y+2}$,
$A=\frac{\sqrt{2}}{2},B=1, C=-\frac{\sqrt{2}}{2}, D=1$.
Integrating between $0$ and $\infty$, we get:
$\frac{\pi}{\sqrt{6}}$.
