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?
 A: 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:
$\frac{\pi}{\sqrt{6}}$.
A: $$I=\int_0^1\sqrt{\sqrt{\frac{4}{x}-3}-1}dx$$
Start with integration by parts
$$u=\sqrt{\sqrt{\frac{4}{x}-3}-1}⇒du=d\left(\sqrt{\sqrt{\frac{4}{x}-3}-1}\right)$$
$$dv=dx⇒v=x$$
$$I=\int_0^1xd\left(\sqrt{\sqrt{\frac{4}{x}-3}-1}\right)$$
Substitute $u=\sqrt{\sqrt{\frac{4}{x}-3}-1}$
$$I=\int_0^\infty\frac{4du}{u^4+2u^2+4}$$
Substitute $u→\sqrt2u $
$$I=\int_0^\infty\frac{4\sqrt2dx}{4u^4+4u^2+4}=\int_{0}^\infty\frac{\sqrt2du}{u^4+u^2+1}$$
I'll let WolframAlpha take it from here
A: We can rewrite the integrand as a rational function with just a few substitutions:
$$\begin{align*}
I &= \int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}} \, dx \\[1ex]
&= 8 \int_0^1 \sqrt x \sqrt{1-x} \, \frac{dx}{(3x^2+1)^2} \tag{1} \\[1ex]
&= 16 \int_0^\infty \frac{x^2 (x^2+1)}{\left(x^4+2x^2+4\right)^2} \, dx \tag{2}
\end{align*}$$

*

*$(1)$ : substitute $x\mapsto\frac{4x^2}{3x^2+1}$

*$(2)$ : substitute $x\mapsto\sqrt{\frac{1-x}x}$
and this could probably be made even simpler. But we can evaluate the current form quite readily with complex analysis.
Let $\mathcal I=2I$ be the same integral but taken over the entire real line. Replace $x$ with complex $z$ and integrate along a semicircular contour with radius $R>\sqrt2$. It's easy to see the integral over the circular arc will vanish as $R\to\infty$, and all we need to do is add up the residues at the order-$2$ poles $z=\frac{\pm1+i\sqrt3}{\sqrt2}$.
