TITLE: Expression for $\int_0^1 x^n(1-x)^{n}/(1+x^2) \ dx$
QUESTION [10 upvotes]: An answer to this question makes clever use of an integral of this form:
$$\int_0^1 \frac{x^n(1-x)^n}{1+x^2} dx$$
Is there a closed form for this for arbitrary positive integer $n$?
(I expect this question has been asked before, but I couldn't find it. So perfectly happy if you can find it and close this question out.)

REPLY [9 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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With $\ds{0 < z < 1}$:

\begin{align}&\sum_{n\ =\ 0}^{\infty}z^{n}
\int_{0}^{1}{x^{n}\pars{1 - x}^{n} \over 1 + x^{2}}\,\dd x
=\int_{0}^{1}{1 \over 1 + x^{2}}\,{1 \over 1 - zx\pars{1 - x}}\,\dd x
\\[5mm]&={1 \over z}\int_{0}^{1}
{\dd x \over \pars{1 + x^{2}}\pars{x^{2} - x + z^{-1}}}
\\[5mm]&={8\root{z}\pars{3z - 2}\,{\rm arccsc}\pars{2/\root{z}}
-\root{4 - z}\braces{\bracks{\pi - 2\ln\pars{2}}z - \pi} \over 4\root{4 - z}\pars{2z^{2} - 2z + 1}}
\end{align}

\begin{align}
&\color{#66f}{\large\int_{0}^{1}{x^{n}\pars{1 - x}^{n} \over 1 + x^{2}}\,\dd x}
\\[5mm]&=\bracks{z^{n}}\pars{{8\root{z}\pars{3z - 2}\,{\rm arccsc}\pars{2/\root{z}}
-\root{4 - z}\braces{\bracks{\pi - 2\ln\pars{2}}z - \pi} \over 4\root{4 - z}\pars{2z^{2} - 2z + 1}}}
\end{align}