In an article in American Mathematical Monthly, Don Zagier states the following equality $$\sum\limits_{r=0}^\infty\left(\int_0^1t^r(1-t)^r dt\right)x^r=\int_0^1\frac{dt}{1+xt+xt^2}$$ however I don't understand how he obtains a positive $xt$ on the denominator instead of a negative. The article: https://www.jstor.org/stable/2324560?seq=1.
Here is my working.
Let
\begin{equation*}
S= \sum\limits_{r=0}^\infty\left(\int_0^1t^r(1-t)^r dt\right)x^r=\int_0^1\left(\sum\limits_{r=0}^\infty t^r(1-t)^rx^r\right)dt.
\end{equation*}
Then
\begin{align*}
\sum\limits_{r=0}^\infty t^r(1-t)^rx^r=\sum\limits_{r=0}^\infty(xt(1-t))^r&=1+xt(1-t)+(xt(1-t))^2+\ldots\\
&=\frac{1}{1-xt(1-t)}=\frac{1}{1-xt+xt^2}.
\end{align*}
Hence
\begin{equation*}
S=\int_0^1\frac{dt}{1-xt+xt^2}.
\end{equation*}