Dalzell-type integrals for convergents to $2\pi$ 
Should $\pi$ be harder than $\log(2)$?

Simple integrals relate $\pi$ to its first two convergents $3$ and $\dfrac{22}{7}$.
$$
\begin{align}
\int_{0}^{1}\frac{2x(1-x)^2}{1+x^2}dx&=\pi-3\\
\\
\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx&=\frac{22}{7}-\pi (Dalzell)
\end{align}
$$
However, the following $\dfrac{333}{106}$ and $\dfrac{355}{113}$ are more involved. Is there an integral that proves $\pi > 333/106$?
In contrast, analogous integrals that evaluate to the error between $\log(2)$ and its first four convergents are easy to find.
$$
\begin{align}
\int_0^1\frac{2x}{1+x^2}dx
&=
\log\left(2\right) \\
\\
\int_0^1\frac{(1-x)^2}{1+x^2}dx
&=
1-\log\left(2\right) \\
\\
\int_0^1\frac{x^2(1-x)^2}{1+x^2}dx
&=
\log\left(2\right)-\frac{2}{3} \\
\\
\int_0^1\frac{x^4(1-x)^2}{1+x^2}dx
&=\frac{7}{10}-\log\left(2\right) \\
\end{align}
$$
Proof that $\frac{2}{3} < \log(2) < \frac{7}{10}$

Why not $2\pi$?

Comparing the list of convergents to $\pi$
$$3,\dfrac{22}{7},\dfrac{333}{106},\dfrac{355}{113},...$$
to that of $2\pi$ 
$$6,\dfrac{19}{3},\dfrac{25}{4},\dfrac{44}{7},\dfrac{333}{53},\dfrac{710}{113},...$$
shows that the convergents for $2\pi$ are not simply twice the convergents for $\pi$.
The first four convergents of $\pi$ have their correspondent fractions in the list for $2\pi$, but two more approximations appear: $$\pi \approx \frac{19}{6}$$  $$\pi\approx\frac{25}{8}$$
The list of integrals for the first four convergents of $2\pi$ is simpler than that of $\pi$.
$$
\begin{align}
4\int_0^1 \dfrac{x(1-x)^2}{1+x^2}dx&=2\pi-6\\
\\
4\int_0^1 \dfrac{x^3(1-x)^2}{1+x^2}dx&=\dfrac{19}{3}-2\pi\\
\\
\dfrac{1}{2} \int_0^1 \dfrac{x(1-x)^4(1+4x+x^2)}{1+x^2}dx&=2\pi-\dfrac{25}{4}\\
\\
2\int_0^1 \dfrac{x^4(1-x)^4}{1+x^2}dx&=\dfrac{44}{7}-2\pi\\
\end{align}
$$
The main question is:

Are there integer $m$,$n$ and a rational $q$ for the third convergent of $2\pi$? $$q\int_0^1 \dfrac{x^m(1-x)^n}{1+x^2}dx=2\pi-\frac{25}{4}$$ If not, how to prove it?

and a more speculative one:

What rational multiple of $\pi$ should we focus on?


It's not a definitive and complete answer at all and it's too lenghty for a comment. It's an empirically based answer.
Let,
$\displaystyle J(m,n)=\int_0^1 \dfrac{x^m(1-x)^n}{1+x^2}dx$ 
You are searching an integer relation between $J(m,n)$ and $2\pi-\dfrac{25}{4}$
Here is a program for PARI-GP to search empirically for $m,n$:
$J(m,n)={intnum(x=0,1,x^m*(1-x)^n/(1+x^2))};$
$scan(r,p)=\{$
$for(m=1,r,for(n=1,r,L=lindep([J(m,n),2*Pi-25/4]);if(abs(L[1])<p\&\&abs(L[2])<p,print(m,\text{" "},n,\text{" "},L[1],\text{" "},L[2]);return)))\}$
r=max(m,n).
p=precision for integers coefficients.
To use:
start to define the precision 
\p 100
(precision to compute integrals=100)
scan(50,1000)
(i have tested this, no m,n found)
If m,n are found,
$L[1]J(m,n)+L[2]\times 2\pi=0$
$L[1],L[2]$ are integers.
If you replace 25/4 by 19/3 or by 44/7 this program finds an integer relation.
Bon courage dans tes recherches !
ADDENDUM:
To search for integer relation between $J(m,n),\pi,1$
i mean searching for integers $a,b,c$ and integers $m,n$ such that:
$aJ(m,n)+b\pi+c=0$
here is a variant of the program above:
$scanb(r,p)=\{$
$for(m=1,r,for(n=1,r,L=lindep([J(m,n),Pi,1]);if(abs(L[1])<p\&\&abs(L[2])<p\&\&abs(L[3])<p,print(m," ",n," ",L[1]," ",L[2]," ",L[3]))))
\}$
