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.

Note 
$$\frac{1}{x\log^2x}=-\frac{d}{dx}\frac{1}{\log x} $$
and hence
\begin{eqnarray}
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\\
&=&-\int_{0}^{1}
(x\log x + 1 - x)\log(1+x)d\frac{1}{\log x}\\
&=&-(x\log x + 1 - x)\log(1+x)\frac{1}{\log x}\bigg|_0^1+\int_{0}^{1}\frac{1}{\log x}d[
(x\log x + 1 - x)\log(1+x)]\\
&=&\int_0^1\frac{1}{\log x}\left(\frac{x \log x+1-x}{x+1}+\log x \log (x+1)\right)dx\\
&=&\int_0^1\log(x+1)dx+\int_0^1\frac{x}{x+1}+\int_0^1\frac{1}{\log x}\frac{1-x}{x+1}dx\\
&=&\log 2+J,
\end{eqnarray}
where
$$ J=\int_0^1\frac{1}{\log x}\frac{1-x}{x+1}dx.$$
Define
$$ f(a)=\int_0^1x^a\frac{1-x}{x+1}dx. $$
Then
\begin{eqnarray}
f(a)&=&\int_0^1\sum_{n=0}^\infty(-1)^nx^{a+n}(1-x)dx
&=&\sum_{n=0}^\infty(-1)^n\left(\frac{1}{n+a+1}-\frac{1}{n+a+2}\right)
\end{eqnarray}
and hence
$$ \int_0^a f(a)da=\sum_{n=0}^\infty(-1)^n\log\frac{n+a+1}{n+a+2},$$
and
$$ J=\lim_{a\to0}\sum_{n=0}^\infty(-1)^n\log\frac{n+a+1}{n+a+2}=\sum_{n=0}^\infty(-1)^n\log\frac{n+1}{n+2}=\log\frac{2}{\pi}.$$
Thus
$$ I=\log\frac{4}{\pi}. $$
