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.
 A: $ \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*{} $
A: To answer your question on evaluating $I_1$, partial fraction decomposition yields a rather tedious result to work with. Instead, there is a nice substitution one can make by taking advantage of the symmetry within the integral. Consider the general case
$$I\equiv\int\limits_0^{+\infty}\frac {\mathrm dt}{t^4+\alpha t^2+\beta^2}=\int\limits_0^{+\infty}\frac 1{t^2}\frac {\mathrm dt}{\left(t-\frac {\beta}t\right)^2+\alpha+2\beta}$$
Making the inverse substitution $t\mapsto\tfrac {\beta}t$ on the right-hand side, then we get that
$$\beta I=\int\limits_0^{+\infty}\frac {\beta}{t^2}\frac {\mathrm dt}{\left(t-\frac {\beta}t\right)^2+\alpha+2\beta}=\int\limits_0^{+\infty}\frac {\mathrm dt}{\left(t-\frac {\beta}t\right)^2+\alpha+2\beta}$$
Adding the two integrals together, then the left-hand side becomes $2\beta I$
\begin{align*}
I & =\frac 1{2\beta}\int\limits_0^{+\infty}\left(1+\frac {\beta}{t^2}\right)\frac {\mathrm dt}{\left(t-\frac {\beta}t\right)^2+\alpha+2\beta}\\ & =\frac 1{2\beta}\int\limits_{-\infty}^{+\infty}\frac {\mathrm dx}{x^2+\alpha+2\beta}\\ & =\frac {\pi}{2\beta\sqrt{\alpha+2\beta}}
\end{align*}
Where the substitution $x=t-\tfrac {\beta}t$ was made to get to the penultimate step. Your integral $I_1$ is the case when $\alpha=\beta=2$.

Another way of evaluating your original integral would be to start off directly with the expression
$$I\equiv\int\limits_0^{+\infty}\frac {t^2(t^2+1)}{(t^4+at^2+b^2)^2}\,\mathrm dt=\int\limits_0^{+\infty}\frac {t^2+1}{t^2}\frac {\mathrm dt}{\left[\left(t-\frac bt\right)^2+a+2b\right]^2}$$
In a similar fashion, enforce the substitution $t\mapsto\tfrac bt$ and clear the resulting nested fractions to obtain
$$bI=\int\limits_0^{+\infty}\frac {b(t^2+1)}{t^2}\frac {\mathrm dt}{\left[\left(t-\frac bt\right)^2+a+2b\right]^2}=\int\limits_0^{+\infty}\frac {t^2+b^2}{t^2}\frac {\mathrm dt}{\left[\left(t-\frac bt\right)^2+a+2b\right]}$$
Adding the two integrals together and observing that the numerator can be factored as $(b+1)(t^2+b)$ gives
\begin{align*}
I & =\frac {b+1}{2b}\int\limits_0^{+\infty}\left(1+\frac b{t^2}\right)\frac {\mathrm dt}{\left[\left(t-\frac bt\right)^2+a+2b\right]^2}\\ & =\frac {b+1}{2b}\int\limits_{-\infty}^{+\infty}\frac {\mathrm dx}{(x^2+a+2b)^2}\\ & =\frac {\pi(b+1)}{4b(a+2b)\sqrt{a+2b}}
\end{align*}
Where the result proved in the first-half of this answer was used. Substituting $a=b=2$ immediately gives the answer as
$$\int\limits_0^{+\infty}\frac {t^2(t^2+1)}{(t^4+2t^2+4)^2}\,\mathrm dt\color{blue}{=\frac {\pi}{16\sqrt 6}}$$
Your original integral is $16$ times the result in blue.
A: *

*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}

*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
$$I= \int_0^1 y(x)dx =\int_0^\infty x(y)dy
= \int_0^\infty \frac{4}{y^4+2y^2+4} dy=\frac\pi{\sqrt6}
$$
