Evaluate the integral $I=\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx$ To evaluate the integral
$$I=\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx$$
I define $t=-1+\sqrt{\frac{4}{x}-3}$, then we have $x=\frac{4}{(t+1)^2+3}$ and
$$dx=\frac{-8(t+1)}{(t^2+2t+4)^2}dt$$
So we get
$$\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx=8\int_0^\infty \frac{\sqrt{t}(t+1)}{(t^2+2t+4)^2}\ dt$$
Again, let $s=\sqrt{t}$, then $dt=2sds$ and the integral becomes
$$8\int_0^\infty \frac{s(s^2+1) 2s}{(s^4+2s^2+4)^2}\ ds=16\int_0^\infty \frac{s^2(s^2+1)}{(s^4+2s^2+4)^2}\ ds$$
Next, I write the integrand as
$$\frac{s^2(s^2+1)}{(s^4+2s^2+4)^2}=\frac{1}{s^4+2s^2+4}-\frac{s^2+4}{(s^4+2s^2+4)^2}$$
so I have to evaluate
$$I_1=\int_0^\infty \frac{1}{s^4+2s^2+4}\ ds\  and\ I_2=\int_0^\infty \frac{s^2+4}{(s^4+2s^2+4)^2}\ ds$$
But when I evaluate the integral $I_1$, I get a weird result as follows:
$$I_1=\int_0^\infty \frac{1}{(s^2+2)^2-(\sqrt{2}s)^2}\ ds
=\int_0^\infty \frac{1}{(s^2+\sqrt{2}s+2)(s^2-\sqrt{2}s+2)}\ ds=\int_0^\infty \left(\frac{\frac{1}{4\sqrt{2}}s+\frac{1}{4}}{s^2+\sqrt{2}s+2}+\frac{\frac{-1}{4\sqrt{2}}s+\frac{1}{4}}{s^2-\sqrt{2}s+2}\right)\ ds$$
and if we separate two parts and evaluate the integrals, we would get two divergent improper integrals. So how can I find $I_1$?(Hope that the method is elementary and without complex analysis.)
Another question: I'm a beginner at learning complex analysis. I conjecture that we can evaluate the integral $I$ in complex analysis (or maybe not worked). Hope everybody can give me some hints or solutions with the method in complex analysis.

$ \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{2}{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_{1}^{\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*{} $
