<s>[INST] Asymptotic quality of rational approximations to $\pi$ Dalzell's integral
$$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi$$
is case $n=2$ of the generalization
$$\int_0^1 \frac{x^{n+2}(1-x)^{2n}}{2^{n-2}(1+x^2)}dx = \frac{p_n}{q_n}-\pi$$
Such an integral gives rational approximations to $\pi$ from above: $4$, $\dfrac{19}{6}$, $\dfrac{22}{7}$, $\dfrac{377}{120}$,...
The qualities M of the latter three fractions are $2.057, 3.429, 1.98669,$ according to the definition
$$\Bigg \|{\pi-\frac{p_n}{q_n}}\Bigg \|=\frac{1}{q^{M_n}}$$

How is the asymptotic quality of this approximation sequence $\displaystyle\lim_{n \to \infty} M_n$  computed?

 [/INST] This answer combines @AntonioVargas and @GregMartin results.
Let us start from the approximation in Laplace's method (https://en.wikipedia.org/wiki/Laplace%27s_method)
$$\int_a^b h(x)e^{Mg(x)}dx \approx \sqrt{\frac{2\pi}{M\|g''(x_0)\|}}h(x_0)e^{Mg(x_0)}$$
and rewrite the general integral in the question as
$$\int_0^1 \frac{4x^2}{1+x^2}e^{n\log\left(\dfrac{x(1-x)^2}{2}\right)} dx$$
so we may identify
$$h(x)=\dfrac{4x^2}{1+x^2}$$
$$g(x)=\log\left(\dfrac{x(1-x)^2}{2}\right)$$
$$g'(x)=\dfrac{1-3x}{x(1-x)}$$
$$g''(x)=-\dfrac{1-2x+3x^2}{x^2(1-x)^2}$$
The position of the unique global maximum of $g(x)$ is obtained from $g'(x_0)=0$
$$x_0=\frac{1}{3}$$
Substituting into $h(x)$, $g(x)$ and $g''(x)$, we obtain
$$h(x_0)=h\left(\dfrac{1}{3}\right)=\frac{4\left(\dfrac{1}{3}\right)^2}{1+\left(\dfrac{1}{3}\right)^2}=\frac{2}{5}$$
$$g(x_0)=log\left(\dfrac{x_0(1-x_0)^2}{2}\right) = log\left(\frac{2}{27}\right)$$
$$g''(x_0)=g''\left(\dfrac{1}{3}\right)=-\frac{27}{2}$$
Finally,
$$\sqrt{\frac{2\pi}{M\|g''(x_0)\|}}h(x_0)e^{Mg(x_0)} = \sqrt{\dfrac{2\pi}{n\frac{27}{2}}}\frac{2}{5}e^{n\log\left(\dfrac{2}{27}\right)} = \frac{4}{15}\sqrt{\dfrac{\pi}{3n}}\left(\dfrac{2}{27}\right)^n,$$
so 

$$\int_0^1 \frac{x^{n+2}(1-x)^{2n}}{2^{n-2}(1+x^2)}dx \sim \frac{4}{15}\sqrt{\dfrac{\pi}{3n}}\left(\dfrac{2}{27}\right)^n$$

for large $n$.
The asymptotic quality is therefore
$$\theta = \lim_{n \to \infty} -\frac{\log\left(\dfrac{p_n}{q_n}-\pi\right)}{\log(q_n)} = \lim_{n \to \infty} -\frac{\log\left( \dfrac{4}{15}\sqrt{\dfrac{\pi}{3n}}\left(\dfrac{2}{27}\right)^n \right)}{\log( (2e^3)^n)} = \frac{3log(3)-log(2)}{3+log(2)} \approx 0.7  < 1,$$
which is low.
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