Evaluate the integral $I=\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx$

Question

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.

Answer 1

1. Continue with \begin{align} I=&\int_0^1 \sqrt{-1+\sqrt{\frac{4}{x}-3}}\ dx\\ =&\ 8\int_0^\infty \frac{\sqrt{t}(t+1)}{\left(t^2+2t+4\right)^2}dt\overset{x=\sqrt{\frac t2}} =2\sqrt2\int_0^\infty \frac{2+\frac1{x^2}}{\left(x^2+\frac1{x^2}+1\right)^2}\overset{x\to\frac1x}{dx}\\ =& \ 3\sqrt2 \int_0^\infty \frac{1+\frac1{x^2}}{\left(x^2+\frac1{x^2}+1\right)^2}dx\overset{x-\frac1x\to x } =3\sqrt2 \int_{-\infty}^\infty \frac{1}{\left(x^2+3\right)^2}dx= \frac\pi{\sqrt6} \end{align}
2. Alternatively, let $$y= \sqrt{-1+\sqrt{\frac{4}{x}-3}}\implies x=\frac{4}{y^4+2y^2+4}$$ and it is much simpler integrating in $y$ instead $$I= \int_0^1 y(x)dx =\int_0^\infty x(y)dy = \int_0^\infty \frac{4}{y^4+2y^2+4} dy=\frac\pi{\sqrt6} $$