Why does this lead to zero when integrating a non-negative function?

Question

When I try to integrate this $\int_0^{1} \frac{\log ^2(x+1) \log ^2(1-x)}{x} \, dx$, it leads to 0.

And when I use NIntegrate, it turns into a positive number, which seems to be right.

The codes are shown below.

Clear["Global`*"];
Plot[Log[1 + x]^2 Log[1 - x]^2/x, {x, 0, 1}]
Integrate[Log[1 + x]^2 Log[1 - x]^2/x, {x, 0, 1}]
NIntegrate[Log[1 + x]^2 Log[1 - x]^2/x, {x, 0, 1}]

Answer 1

I cannot answer why, but I can provide an analytical solution with MA. We have

$$\int_0^1 \frac{\log ^2(1-x) \log ^2(x+1)}{x} \, dx= \frac{1}{12} \int_0^1 \frac{\log ^4\left(1-x^2\right)}{x} \, dx +\frac{1}{12} \int_0^1 \frac{\log ^4\left(\frac{1-x}{x+1}\right)}{x} \, dx -\frac{1}{6} \int_0^1 \frac{\log ^4(1-x)}{x} \, dx -\frac{1}{6} \int_0^1 \frac{\log ^4(x+1)}{x} \, dx.$$

Each of them can be correctly computed with MA

i1 = 1/12 Integrate[Log[1 - x^2]^4/x, {x, 0, 1}]
(*Zeta[5]*)
i2 = 1/12 Integrate[Log[(1 - x)/(1 + x)]^4/x, {x, 0, 1}]
(*(31 Zeta[5])/8*)
i3 = -1/6 Integrate[Log[1 - x]^4/x, {x, 0, 1}]
(*-4 Zeta[5]*)
i4 = -1/6 Integrate[Log[1 + x]^4/x, {x, 0, 1}]
(*1/6 (-(2/3) \[Pi]^2 Log[2]^3 + (4 Log[2]^5)/5 + 21/2 Log[2]^2 Zeta[3] +
   24 (Log[2] PolyLog[4, 1/2] + PolyLog[5, 1/2] - Zeta[5]))*)

Combining together we obtain

i = i1 + i2 + i3 + i4

$$4 \text{Li}_5\left(\frac{1}{2}\right)+\text{Li}_4\left(\frac{1}{2}\right) \log (16)-\frac{25 \zeta (5)}{8}+\frac{7}{4} \zeta (3) \log ^2(2)+\frac{2 \log ^5(2)}{15}-\frac{1}{9} \pi ^2 \log ^3(2).$$