How to find $\int_0^1\!\! \left(\frac1{\log^2\left(1-x\right)}-\frac1{x^2}+\frac1x-\frac1{12}\right) \frac{\mathrm{d}x}x$

How to show that

$$\int_0^1\!\! \left(\frac1{\log^2\left(1-x\right)}-\frac1{x^2}+\frac1x-\frac1{12}\right) \frac{\mathrm{d}x}x=\frac{\ln{(2\pi)}}{12}-\frac{5}{24}+\frac{1}{2\pi^2}\sum_{n=1}^\infty \frac{\ln{n}}{n^2}$$ ?

I have no any idea.

• Is it just $\dfrac{1}{\log^2 (1-x)}$? It won't make sense when $x > 1$. – r9m Dec 2 '15 at 22:31
• As an aside, $~\displaystyle\sum_{n=1}^\infty\frac{\ln n}{n^2}=-\zeta'(2).$ – Lucian Dec 2 '15 at 23:54
• Similarly to what I wrote in my very first comment to this question, your integral is nothing more than a rigorous way of asking for a regularization of $$\displaystyle\int_0^1\ln^k(1-x)~x^{n-1}~dx~=~\frac{d^k}{dm^k}~\int_0^1(1-x)^{m-1}~x^{n-1}~dx~=~\frac{d^k}{dm^k}~B(m,n)$$ to negative values of n, where B represents Euler's famous beta function. Also, since k is negative as well, we are dealing with integration rather than differentiation. – Lucian Oct 3 '16 at 5:18