According to 8.111 from Lewin's book "Polylogarithms and associated functions", it is expected that $$\int\limits_0^2\frac{\ln{(1-x)}\ln{(1+x)}}{x}\,dx=Li_3(-3)+\zeta(3)-2Li_3(3)+$$ $$\ln{3}\left[Li_2(-1)-Li_2(2)\right]+i\pi\left[Li_2(3)-Li_2(-1)-2\zeta(2)\right ], \tag{1} $$ as Lewin clarifies in his another book "for some choice of branches of $Li_3$, $Li_2$ and $\ln$" ("Structural properties of polylogarithms", p. 144). However, it seems http://www.wolframalpha.com (and naive application of some polylogarithm functional identities) doesn't confirm (1) and requires an extra term in the imaginary part of the r.h.s., namely $-i\pi\ln^2(3)$. From where this extra term comes from?

According to 8.111 from Lewin's book "Polylogarithms and associated functions", it is expected that $$\int\limits_0^2\frac{\ln{(1-x)}\ln{(1+x)}}{x}\,dx=Li_3(-3)+\zeta(3)-2Li_3(3)+$$ $$\ln{3}\left[Li_2(-1)-Li_2(2)\right]+i\pi\left[Li_2(3)-Li_2(-1)-2\zeta(2)\right ], \tag{1} $$ as Lewin clarifies in his another book "for some choice of branches of $Li_3$, $Li_2$ and $\ln$" ("Structural properties of polylogarithms", p. 144). However, it seems numerical computation (at http://www.wolframalpha.com), as well as naive application of some polylogarithmic functional identities to the r.h.s of (1), doesn't confirm (1) and requires an extra term in the imaginary part of the r.h.s., namely $-i\pi\ln^2(3)$. From where this extra term comes from?