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May I asked, is this integral is an obvious that the result is zero?

I can't see how this is zero.

$$\int_{0}^{\infty}{\ln^{2n}(x)\over 1-x^2}\mathrm dx=0\tag1$$ $n\ge 1$

Can anyone show it, I just need to know.

We can split

$${1\over 2}\int_{0}^{\infty}\ln^{2n}(x)\left({1\over 1-x}+{1\over 1+x}\right)\mathrm dx\tag2$$

Any hints?

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    Have you tried to split this integral into two new ones, the first with 0 and 1 as bounds, the second one with 1 and $\infty $, before performing a change of variables $ t=1/x $? – 2017-01-26

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First note that the function being integrated is positive on $(0,1)$ and negative on $(1,\infty)$ so it suffices to show $$ \int_0^1\frac{\ln^{2n}(x)}{1-x^2}dx=-\int_1^\infty\frac{\ln^{2n}(x)}{1-x^2}dx $$

Think about what happens to the first integral when you make the $u$ substitution $u=\tfrac{1}{x}$.