<s>[INST] Source and context of $\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 dx/(1+x^2)$? Possibly the most striking proof of Archimedes's inequality $\pi < 22/7$ is an integral formula for the difference:
$$
\frac{22}{7} - \pi = \int_0^1 (x-x^2)^4 \frac{dx}{1+x^2},
$$
where the integrand is manifestly positive.  This formula is "well-known" but its origin remains somewhat mysterious.  I ask:

Who discovered this integral, and in what context?

The earliest reference I know of is Problem A-1 on the 29th Putnam Exam (1968).  According to J.H.McKay's report in the American Math. Monthly (Vol.76 (1969) #8, 909-915), the Questions Committee consisted of N.D.Kazarinoff, Leo Moser, and Albert Wilansky.  Is one of them the discoverer, and if so which one?
The printed solution, both in the Monthly article and in the book by Klosinski, Alexanderson, and Larson, says only "The standard approach, from elementary calculus, applies. By division, rewrite the integrand as a polynomial plus a rational function with numerator of degree less than 2. The solution follows easily."  But surely there's more to be said, because this integral is a minor miracle of mathematics:
$\bullet$ Not only is the integrand manifestly positive, but it is always small: $x-x^2 \in [0,1/4]$ for $x \in [0,1]$, and the denominator $1+x^2$ is at least 1, so $(x-x^2)^4/(x^2+1) < 1/4^4 = 1/256$.  A better upper bound on the integral is $\int_0^1 (x-x^2)^4 dx$, which comes to $1/630$ either by direct expansion or by recognizing the Beta integral $B(5,5)=4!^2/9!$.  Hence $\frac{22}{7} - \pi < 1/630$, which also yields Archimedes's lower bound $\pi > 3\frac{10}{71}$.
$\bullet$ The "standard approach" explains how to evaluate the integral, but not why the answer is so simple.  When we expand
$$
\frac{(x-x^2)^4}{1+x^2} = x^6 - 4x^5 + 5x^4 - 4x^2 + 4 - \frac4{x^2+1},
$$
the coefficient of $x/(x^2+1)$ vanishes, so there's no $\log 2$ term in the integral.  [This much I can understand: the numerator $(x-x^2)^4$ takes the same value $(1\pm i)^4 = -4$ at both roots of the denominator $x^2+1$.]  When we integrate the polynomial part, we might
expect to combine fractions with denominators of 2, 3, 4, 5, 6, and 7, obtaining a complicated rational number.  But only 7 appears: there's no $x$ or $x^3$ term; the $x^4$ coefficient 5 kills the denominator of 5; and the terms $-4x^5-4x^2$ might have contributed denominators of 6 and 3 combine to yield the integer $-2$.
Compare this with the next such integrals
$$
\int_0^1 (x-x^2)^6 \frac{dx}{1+x^2} = \frac{38429}{13860} - 4 \log 2
$$
and
$$
\int_0^1 (x-x^2)^8 \frac{dx}{1+x^2} = 4\pi - \frac{188684}{15015},
$$
which yield better but much more complicated approximations to $\log 2$ and $\pi$...
This suggests a refinement of the "in what context" part of the question:

Does that integral for $(22/7)-\pi$ generalize to give further approximations to $\pi$ (or $\log 2$ or similar constants) that are useful for the study of Diophantine properties of $\pi$ (or $\log 2$ etc.)?

 [/INST] This integral has a series counterpart
$$\sum_{k=0}^\infty \frac{240}{(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)}=\frac{22}{7}-\pi$$
https://math.stackexchange.com/a/1657416/134791
(UPDATE Peter Bala New series for old functions https://oeis.org/A002117/a002117.pdf, 2009, formula 5.1)
Equivalently,
$$\sum_{k=1}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{22}{7}-\pi$$
which may be seen as the first truncation of
$$\sum_{k=0}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{10}{3}-\pi$$
Therefore, this series shows the following path to $\frac{22}{7}$ 
$$\frac{10}{3}-\frac{240}{1·2·3·5·6·7}=\frac{10}{3}-\frac{2·5!}{\frac{7!}{4}}=\frac{10}{3}-\frac{8·5!}{5!6·7}=\frac{1}{3}\left(10-\frac{4}{7}\right)=\frac{1}{3}·\frac{66}{7}=\frac{22}{7}$$
also illustrating how only $7$ remains.
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