Differentiation under the integral sign and a sum
We will see how differentiation under the integral sign, which Mr. Feynman loved, can be applied to derive an otherwise difficult integral. The integral can also be expanded as a Taylor series, thus obtaining an infinite sum.
So, this is the integral:
\begin{equation*}
\displaystyle I(a)=\int_0^1 \, \frac{\ln{(1+a\,(x-x^2))}}{x-x^2}\, dx
\end{equation*}
Applying the differentiation under the integral sign to this:
\begin{align*}
\displaystyle
\frac{\partial}{\partial a} I(a) &= \int_0^1\, \frac{1}{1+a\,(x-x^2)}\, dx\\
&=\frac{\log\left(\frac{1}{2} \, a + \frac{1}{2} \, \sqrt{a^{2} + 4 \, a} + 1\right)}{\sqrt{a^{2} + 4 \, a}} - \frac{\log\left(\frac{1}{2} \, a - \frac{1}{2} \, \sqrt{a^{2} + 4 \, a} + 1\right)}{\sqrt{a^{2} + 4 \, a}}
\end{align*}
Integrating w.r.t \(a\) gives us (answer by friCAS)
\begin{equation*}
\displaystyle I(a)=\log\left( \frac{a+2 -\sqrt{a^{2} + 4 \, a}}{2}\right)^{2} + C
\end{equation*}
Putting \(a=0\) in the above gives \(I(0)=0+C\) and original integral is also 0. Hence, \(C=0\) and the final answer to the integral is:
\begin{equation*}
\displaystyle I(a)=\log\left( \frac{a+2 -\sqrt{a^{2} + 4 \, a}}{2}\right)^{2}
\end{equation*}
Now, we may also expand the integral as a taylor series:
\begin{align*}
\displaystyle
I(a)&=\int_0^1 \, \sum_{n\ge 1}\, \frac{(-1)^{n-1}\, a^{n} (x-x^2)^n}{n\, (x-x^2)} \, dx\\
&=\int_0^1 \, \sum_{n\ge 1}\, \frac{(-1)^{n-1}\, a^{i} (x-x^2)^{n-1}}{n} \, dx\\
&=\int_0^1 \, \sum_{n\ge 0}\, \frac{(-1)^{n}\, a^{n+1} (x-x^2)^{n}}{n+1} \, dx\\
&=\sum_{n\ge 0}\, \frac{(-1)^{n}\, a^{n+1} B(n+1,n+1)}{n+1}\\
&=\sum_{n\ge 0}\, \frac{(-1)^{n}\, a^{n+1} \Gamma\left(n+1\right)\Gamma\left(n+1\right)}{(n+1)\,\Gamma\left(2\, n+1\right)}\\
&=\sum_{n\ge 0}\, \frac{(-1)^{n}\, a^{n+1} \, n!^2}{(n+1)\,(2\, n+1)\, (2n)!}\\
&=\sum_{n\ge 0}\, \frac{(-1)^{n}\, a^{n+1}}{(n+1)\,(2\, n+1)\, \binom{2n}{n}}
\end{align*}
Therefore, we have the following sum!
\begin{equation*}
\displaystyle \sum_{n\ge 0}\, \frac{(-1)^{n}\, a^{n+1}}{(n+1)\,(2\, n+1)\, \binom{2n}{n}}=\log\left( \frac{a+2 -\sqrt{a^{2} + 4 \, a}}{2}\right)^{2}
\end{equation*}
This has got real values for \(a>-4\)
E.g. when \(a=-2\)
\begin{equation*}
\displaystyle \sum_{n\ge 0}\, \frac{2^{n+1}}{(n+1)\,(2\, n+1)\, \binom{2n}{n}}=\frac{\pi^2}{4}
\end{equation*}
and when \(a=1/2\)
\begin{equation*}
\displaystyle \sum_{n\ge 0}\, \frac{(-1)^{n}}{(n+1)\,(2\, n+1)\, \binom{2n}{n}\, 2^{n+1}}=\log\left(2\right)^{2}
\end{equation*}