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 $2\pi+e\approx 9$?