Compute the integral in a closed form : $\int_0^{1}\operatorname{Li}_2(1-x)dx$ How I can find the closed form of the following integration : 
$I=\int_0^{1}\operatorname{Li}_2(1-x)dx$
$J=\int_0^{1}\operatorname{Li}_2(1-x)\operatorname{Li}_2(1-\frac{1}{x})dx$
$K=\int_0^{1}\ln(x)\operatorname{Li}_2(1-\frac{1}{x})dx$
Wolfram alpha give me for $I=$ see here
For $J=$ see here
For $K=$ see here
My attempt first integral is : 
$\operatorname{Li}_2(1-x)=\sum_{n=1}^{\infty}\frac{(1-x)^n}{n^2}$ 
So : 
$I=\sum_{n=1}^{\infty}\frac{\int_0^{1}(1-x)^{n}dx}{n^2}$ 
$=\sum_{n=1}^{\infty}\frac{\beta(1,n+1)}{n^2}$ 
$=\sum_{n=1}^{\infty}\frac{1}{(n+1)n^2}=\zeta (2)-1$ 
But what about from second and last integral ? 
We can use same method ? 
 A: The integrals $J$ and $K$ can be computed using the polylogarithm identity
$$ \operatorname{Li}_2 \left(1 - \frac{1}{x}\right) = - \operatorname{Li}_2(1-x) - \frac{1}{2} \log^2(x) \, , \, x > 0 \, , \tag{1} $$
and the integrals
$$ \int \limits_0^1 \frac{[-\log(x)]^n}{1-x} \, \mathrm{d} x = n! \zeta(n+1) \, , \, n \in \mathbb{N} \, . \tag{2}$$
We have
\begin{align}
K &\stackrel{(1)}{=} \int \limits_0^1 \left[\frac{- \log^3(x)}{2} - \log(x) \operatorname{Li}_2(1-x)\right]\, \mathrm{d} x \\
&= 3 + \left[\left(x - 1 - x \log(x)\right)\operatorname{Li}_2(1-x)\right]_{x = 0}^{x=1}  + \int \limits_0^1 \frac{[1 - x + x \log(x)] \log(x)}{1-x} \, \mathrm{d} x \\
&= 3 + \zeta(2) - 1 - 2 + \int \limits_0^1 \frac{\log^2(x)}{1-x} \, \mathrm{d} x \stackrel{(2)}{=} \zeta(2) + 2 \zeta(3) = \frac{\pi^2}{6} + 2\zeta(3) \, .
\end{align}
Similarly, we can write
$$ - J \stackrel{(1)}{=} \int \limits_0^1 \operatorname{Li}_2^2(1-x) \, \mathrm{d} x + \frac{1}{2} \int \limits_0^1 \log^2(x) \operatorname{Li}_2(1-x) \, \mathrm{d} x \equiv M + \frac{1}{2} N \, .$$
The remaining integrals can be evaluated using integration by parts:
\begin{align}
M &= \zeta^2(2) - 2 \int \limits_0^1 - \log(x) \operatorname{Li}_2(1-x) \, \mathrm{d} x \\
&= \zeta^2(2) - 2 \zeta(2) + 2 \int \limits_0^1 \frac{\log(x)[x - 1 - x \log(x)]}{1-x} \, \mathrm{d} x \\
&= -\zeta(2)(2-\zeta(2)) + 2 + 4 - 2 \int \limits_0^1 \frac{\log^2(x)}{1-x} \, \mathrm{d} x \stackrel{(2)}{=} 6 - \zeta(2)(2-\zeta(2)) - 4 \zeta(3)
\end{align}
and
\begin{align}
N &= \int \limits_0^1 \frac{- x \log(x) [\log^2(x) - 2 \log(x) + 2]}{1-x} \, \mathrm{d} x \stackrel{(2)}{=} 6 \zeta(4) + 4 \zeta(3) + 2 \zeta(2) - 12 \, . 
\end{align}
Therefore,
$$ - J = 3 \zeta(4) - 2 \zeta(3) + \zeta(2)(\zeta(2)-1) = \frac{11 \pi^4}{180} - \frac{\pi^2}{6}  - 2 \zeta(3) \, . $$
