If $A=\int_{0}^{1} x^n(1-x)^ndx$ then $A^{-1} \in \mathbb{N}$.

Expanding on my comment: \begin{align}\int_0^1x^n(1-x)^n\,dx&=\frac{x^{n+1}}{n+1}(1-x)^n\big|^{1}_{0}+\frac{n}{n+1}\int_0^1x^{n+1}(1-x)^{n-1}\,dx\\ &=\frac{n}{n+1}\int_0^1x^{n+1}(1-x)^{n-1}\,dx \end{align} now you repeat the process $n$ times so to get $$\int_0^1x^n(1-x)^n\,dx=\frac{n\cdot (n-1)\cdots 1}{(n+1)\cdot (n+2) \cdots 2 n}\int_0^1x^{2n}\,dx=\frac{n\cdot (n-1)\cdots 1}{(n+1)\cdot (n+2) \cdots 2n\cdot (2n+1)}$$ thus $$\int_0^1x^n(1-x)^n\,dx=\frac{n!n!}{(2n+1)!}$$ you shall conclude.

There is also a way to interpret this integral as a probability and therefore obtain a closed formula instantly. Imagine choosing $n$ blue, $n$ red and one black point uniformly at random on $[0,1]$. What is the probability that there will be $n$ red, one black and $n$ blue points appearing on this segment in this order? On one hand it is our integral - if $x$ is the coordinate of the black point, $x^n(1-x)^n$ is the probability that $n$ red points will be on the left and $n$ blue points on the right. On the other hand all possible arrangements are equally likely - there are $(2n+1)!$ of them in total and $(n!)^2$ of them result in the required order.
It also generalizes nicely and allows to compute values of a Beta function for all positive integer values.

You can use this to prove $$A^{-1}=\frac{(2n+1)!}{n!^2}=(2n+1)\binom{2n}{n}.$$