A note about Case $1$ with $\,a:= 2^{-\frac{2}{3}}\,$ and $\,\displaystyle x\to 2^{\frac{1}{3}}\,$ .
Left side $\,=0\,$ for $\,\displaystyle x=2^{\frac{1}{3}}\,$ because of $\, ax^2-1=0\,$ and $\,\displaystyle ax^2-a^2x+1=\frac{3}{2}\ne 0\,$ .
Right side $\,=0\,$ for $\,\displaystyle x\to 2^{\frac{1}{3}}>1\,$:
$\,\displaystyle \lim_{z\to +0 \\x>0} z^x\ln z=0\,$ and therefore $\,\displaystyle \left(\frac{a^2x^2-x+a}{ax}\right)^x \ln \frac{a^2x^2-x+a}{ax} \to 0\,$ for $\,\displaystyle x\to 2^{\frac{1}{3}}>0\,$
$\,\displaystyle \left(\frac{a^2x^2-x+a}{ax}\right)^x \frac{a(ax^2-1)}{a^2x^2-x+a} = \left(a^2x^2-x+a\right)^{x-1} (ax^2-1) x^{-x} a^{1-x} = 0\,$
for $\,\displaystyle x=2^{\frac{1}{3}}\,$ because of $\,x-1>0\,$ and $\,\displaystyle a^2x^2-x+a=0\,$ and $\, ax^2-1=0\,$ .
If you like to work with recursions for e.g. $\,a>0,\,a\neq 1\,$
it can make sense to choose $\,\displaystyle z:=x+\frac{1}{ax}\,$ so that you get
$\displaystyle x=f_{1,2}(z):=\frac{z}{2}\pm\sqrt{(\frac{z}{2})^2-\frac{1}{a}}\,$ and $\,\displaystyle y=(z-a)^a-\left(az-\frac{1}{a}\right)^x\,$ .
We get $\,\displaystyle \frac{dy}{dz}=(z-a)^{a-1} - \left(az-\frac{1}{a}\right)^x \left(\frac{x}{z-\frac{1}{a^2}}+\frac{1}{2-\frac{z}{x}}\ln\left(az-\frac{1}{a}\right)\right)$
and a possible recursion with $\displaystyle z_0>\max\left(a;\frac{2}{\sqrt{a}};\frac{1}{a^2}\right)\,$ could be
$\displaystyle z_{n+1}=a+\left(\left(az_n-\frac{1}{a}\right)^{f(z_n)} \left(\frac{f(z_n)}{z_n-\frac{1}{a^2}}+\frac{1}{2-\frac{z_n}{f(z_n)}}\ln\left(az_n-\frac{1}{a}\right)\right)\right)^{\frac{1}{a-1}}\,$
with $\,f(z)\in\{f_1(z);f_2(z)\}\,$ .
For time reasons I haven't checked this recursion, sorry. But it's a try to simplify the calculations.
Note:
Because of $\,\displaystyle \frac{dy}{dx}=\frac{dy}{dz}\frac{dz}{dx}\,$ with $\,\displaystyle \frac{dy}{dx}:=0\,$ we have $\,\displaystyle \frac{dy}{dz}=0\,$ or $\,\displaystyle \frac{dz}{dx}=0\,$ .
$\displaystyle \frac{dz}{dx}=0\,$ means $\,ax^2-1=0\,$ which leads directly to my comment about Case $1$ .
This strengthens the claim that the substitution $\,\displaystyle z:=x+\frac{1}{ax}\,$ makes sense.
Best Answer
Hint : Solve the inequality $$\frac{y}{1+y} \ge \frac{\pi}{4}$$
and check range of $y$.
Putting together, the intimidating integral reduces to $$\int_0^4 \frac{x}{4} \, dx$$