Is there a closed form for $\int_0^1 \binom{1}{x}\frac{\log^2(1-x)}{x}\ \mathrm{d}x$? Do we know if there is a closed form for 
$$
I :=\int_0^1 \binom{1}{x}\frac{\log^2(1-x)}{x}\ \mathrm{d}x\mathrm{?}
$$
Wolfram alpha gives an approximation of $2.66989$ which may be equivalent to:
$$10\sqrt{\frac{2\pi}{77\log(\pi)}}.$$
As stated by @Mariusz Iwaniuk, in the comments, we have the equivalent representation of
$$I\equiv \int_0^1 \frac{\sin(\pi x)\log^2(1-x)}{\pi x^2(1-x)}.$$Another question, presumably simpler, could be 
$$\int_0^1 \binom{1}{x}\frac{\log(1-x)}{x} \equiv -\int_0^1 \binom{1}{x}\frac{\mathrm{Li}_1(x)}{x}.$$ I believe I can find a closed form for the latter; if I do, I will edit the post. In general, I am curious as to if we may be able to somehow employ Ramanujan's Beta integral or any of the other Beta integrals. Another approach may be the series representation for $\binom{1}{x}$. Thanks!
 A: Using Mariusz Iwaniuk's hint of $\binom{1}{x}=\frac{\sin(\pi x)}{\pi x(1-x)}$, we can rewrite $I$ as $$\int_0^1 \frac{\ln^2(1-x) \sin(\pi x)}{\pi x^2(1-x)} dx$$
We can split the integrand up as $$\int_0^1 \frac{\ln^2(1-x)}{x^2} \frac{\sin(\pi x)}{\pi (1-x)} dx$$ 
and then write a Taylor series about $x=1$ for $\frac{\sin(\pi x)}{\pi (1-x)}$ as $$\sum_{k=0}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!} (x-1)^{2k}$$
Plugging this into the integral, I get $$\sum_{k=0}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!} \underbrace{\int_0^1 \frac{\ln^2(1-x)}{x^2} (x-1)^{2k} dx}_{I_k}$$
Other than $I_0 = \frac{\pi^2}{3}$, Mathematica says that $I_k = 2 (\psi^{(1)}(2k) + k\psi^{(2)}(2k)) = \frac{\pi^2}{3} - 2 H_{2k-1}^{(2)}+4k H_{2k-1}^{(3)} - 4k \zeta(3)$
This therefore means that $$I = \frac{\pi^2}{3} + \sum_{k=1}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!}\left(\frac{\pi^2}{3} - 2 H_{2k-1}^{(2)}+4k H_{2k-1}^{(3)} - 4k \zeta(3)\right)$$
Splitting this up into four distinct sums: $$\frac{\pi^2}{3} + \frac{\pi^2}{3}\underbrace{\sum_{k=1}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!}}_{S_1} - 2 \underbrace{\sum_{k=1}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!}H_{2k-1}^{(2)}}_{S_2}+4\underbrace{\sum_{k=1}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!}k H_{2k-1}^{(3)}}_{S_3} - 4\zeta(3)\underbrace{\sum_{k=1}^{\infty} \frac{(-1)^k \pi^{2k}}{(2k+1)!}k}_{S_4} $$
It is easy to see that $S_1 = -1$, by utilizing the series for $\sin(x)$. Also $S_4 = -\frac{1}{2}$, so it reduces to $-2S_2 + 4S_3 + 2\zeta(3)$. This can be made into $2\zeta(3)$ plus a power series of $\pi$, but I was not able to simplify $S_2$ or $S_3$.
