Integral $\int_0^1 \frac{x\log x+1-x}{x \log^2 x}\log(1+x)\, dx=\log\frac{4}{\pi}$ Hi I am trying to prove this
$$
I:=\int_{0}^{1}
{x\log\left(\,x\,\right) + 1 - x \over x\log^{2}\left(\,x\,\right)}\,
\log\left(\,1 + x\,\right)\,{\rm d}x=\log\left(\,4 \over \pi\,\right).
$$
Thanks.
This is just a beautiful integral for many reasons.  Logs are everywhere and an inspirational solution!!!  I am not sure if breaking it up into three separate pieces is of any use, I  tried that by writing
$$
I=\int_0^1\frac{ \log(1+x)}{\log x}dx+\int_0^1\frac{\log(1+x)}{x \log^2 x}dx-\int_0^1\frac{\log(1+x)}{\log^2 x}dx
$$
but wasn't sure how to handle these.  Also  note that
$$
\int_0^1 \frac{x\log x+1-x}{x \log^2 x}dx=1,
$$
in case that happened to come up anywhere along the calculation.

Consider
\begin{align*}
\int_0^1 \frac{x\log{x}+1-x}{x}\, x^a\, \log{(1+x)}\, dx &= \int_0^1 \frac{x\log{x}+1-x}{x}\, x^a\, \sum_{n\ge 1} (-1)^{n+1}\frac{x^n}{n}\, dx\\
&=\sum_{n\ge 1} \int_0^1 \,  (-1)^{n+1} (x\log{x}+1-x)\, \frac{x^{a+n-1}}{n}\, dx\\
&= \sum_{n\ge 1} - \frac{\left(-1\right)^{n+1}}{{\left(a + n + 1\right)}^{2} n} + \frac{\left(-1\right)^{n+1}}{{\left(a + n\right)} n}  -\frac{\left(-1\right)^{n+1}}{{\left(a + n + 1\right)} n}\\
\int_0^1 \frac{x\log{x}+1-x}{x \log{x}}\, x^a\, \log{(1+x)}\, dx &= \sum_{n\ge 1} \frac{\left(-1\right)^{n + 1}}{n} \left(\frac{1}{a + n + 1} + \log\left(\frac{a + n}{a+n+1}\right)\right)\tag{$\int da$}\\
\int_0^1 \frac{x\log{x}+1-x}{x (\log{x})^2}\, x^a\, \log{(1+x)}\, dx &= \sum_{n\ge 1} \frac{\left(-1\right)^{n+1} {\left(a + n\right)} \log\left(\frac{a + n}{a + n + 1}\right) + \left(-1\right)^{n+1}}{n}\tag{$\int da$}\\
\end{align*}
Subst. $a=0$
\begin{align*}
\therefore \int_0^1 \frac{x\log{x}+1-x}{x (\log{x})^2}\, \log{(1+x)}\, dx &= \sum_{n\ge 1} \left(-1\right)^{n+1} \log\left(\frac{ n}{ n + 1}\right) + \frac{\left(-1\right)^{n+1}}{n}\hspace{20pt} \text{(Wallis product and log 2)}\\
&= \log{\left(\frac{2}{\pi}\right)}+\log{2} \\
&= \log{\left(\frac{4}{\pi}\right)}
\end{align*}
