<s>[INST] An integral for $2\pi+e-9$ Motivation
Lucian asked about the almost-integer $2\pi+e\approx9$ in a comment to a partially answered why question about $e\approx H_8$. This is more involved than approximations to $\pi$ and logarithms because two transcendental constants are included, as in $e^\pi-\pi\approx20$.
Tried so far
An answer can be crafted from integrals related to $\pi\approx\frac{22}{7}$ and $e\approx\frac{19}{7}$
$$\int_0^1 \frac{x^4(1-x)^4}{1+x^2}dx =\frac{22}{7}-\pi$$
$$\frac{1}{14}\int_0^1 x^2(1-x)^2e^xdx=e-\frac{19}{7}$$
to obtain

$$\int_0^1 x^2 (1-x)^2 \left(\frac{e^x}{14}-\frac{2 x^2 (1-x)^2}{1+x^2}\right) dx = 2\pi+e-9 $$

The visual representation of this integral provided by WolframAlpha shows that $2\pi+e-9$ is positive and small (the integrand is between $0$ and $0.004$ for $0<x<1$), although this is not immediate from the analytic expression.
Moreover, two maxima appear, instead of the single one that is usual in this type of integrals.
Question

Is there a simpler integral with positive integrand in (0,1) that proves $2\pi+e\approx 9$?

 [/INST] Approximations $e\approx \frac{163}{60} $ and $ \pi \approx \frac{377}{120}$ are related to the integrals
$$\frac{1}{2}\int_0^1 (1-x)^2\left(e^x-1-x-\frac{x^2}{2}\right)dx = e-\frac{163}{60}$$
and
$$\frac{1}{2}\int_0^1 \frac{x^5(1-x)^6}{1+x^2}dx = \frac{377}{120}-\pi.$$
Combining both, we can build
$$\frac{1}{2} \int_0^1 (1 - x)^2 \left(e^x - 1 - x - \frac{x^2}{2} - \frac{2 x^5 (1 - x)^4}{1 + x^2}\right) dx = 2\pi+e-9,$$
which explains the result with nonnegative small integrand and a single maximum in $(0,1)$.
WolframAlpha link
This sets the lower bound
$$9<2\pi+e$$
For an upper bound, we may take the integral from a failed attempt to match $9$
$$\int_0^1 x^4(1-x)^4\left(-\frac{e^x}{24024}+\frac{2}{1+x^2}\right) dx = 9+\frac{4}{1001}-2\pi-e$$
and write
$$2\pi+e<9+\frac{4}{1001}$$
Finally,
$$9<2\pi+e<9+\frac{4}{1001}$$
In short, $2\pi+e$ is close to $9$ because
$\pi\approx\dfrac{377}{120}$
from above,
$e \approx \dfrac{163}{60}$
from below and 
$$2·\frac{377}{120}+\frac{163}{60}=9$$
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