Product of ln?

$\begin{aligned} I & = \int_0^1 \ln x \ln (1-x) \ dx & \small \color{#3D99F6}{\text{By integration by parts.}} \\ & = [(x-1)\ln(1-x) - x]\ln x \ \bigg|_0^1 + \int_0^1 \frac {(1-x)\ln(1-x) + x}x \ dx \\ & = 0 + \int_0^1 \frac {\ln(1-x)}x dx - \int_0^1 \ln(1-x) \ dx + \int_0^1 dx \\ & = -Li_2(1) - \bigg[(x-1)\ln(1-x) - x\bigg]_0^1 + x \ \bigg|_0^1 \\ & = - \zeta (2) + 1 + 1 \\ & = 2 - \frac {\pi^2}6 \end{aligned}$

$\implies \dfrac {\pi^2}6 + I = \boxed{2}$ .

Using the fact that $1/(1-x)=1+x+x^2+x^3+....$ One could integrate and find that $-ln(1-x)=x+x^2/2+x^3/3+...$ Substituting into the integral will give: $\pi^2/6-\sum_{i=1}^\infty \int_0^1 x^k * ln(x)/k dx$ Now if we solve the integral: $\int_0^1 x^k * ln(x) dx$ (it just involves u substitution and integration by parts), it will yield: $1/(k+1)^2$ $\pi^2/6+\sum_{i=1}^\infty (1/(k*(k+1)^2))$ Using partial fraction decomposition: $\pi^2/6+\sum_{i=1}^\infty (1/(k*(k+1)^2))=\pi^2/6+\sum_{i=1}^\infty (1/k-1/(k+1)-1/(k+1)^2)=\pi^2/6+\sum_{i=1}^\infty (1/k-1/(k+1))-(\pi^2/6-1)$ The two $\pi^2/6$ cancels out and the $\sum_{i=1}^\infty (1/k-1/(k+1))$ is a telescoping series that evaluates to 1, Hence the answer is 2.