Integral: $\displaystyle\int_0^\infty\!\frac{\tanh^2 x}{x^2}\,dx$

Question

In this answer, Mr. Noam Shalev - nospoon showed that

$$ I = \int_0^\infty \! \frac{\tanh^2 x}{x^2} \, dx = \frac4{\pi^2} \int_0^1 \! \frac1x \, \log^2 \frac{1-x}{1+x} \, dx $$

with using the following substitution

$$ x = \frac{1-t}{1+t} ,\quad dx = - \frac2{(1+t)^2} \, dt. $$

However, this substitution does not allow me to reach this conclusion. Perhaps this substitution is used elsewhere, but either way I don't know. How can I get the above equation?

c.f. Inverse hyperbolic tangent function has a Taylor series

$$ \operatorname{artanh}x = \frac12 \, \log \frac{1+x}{1-x} = \sum_{n=0}^\infty \frac1{2n+1} \, x^{2n+1} $$

which is valid for $x\in(-1,+1)$.

Answer 1

The linked answer is not making a direct connection between the two integrals by this substitution. Rather, it's the subsequent log-integral that is evaluated with its help:

$$\int_0^1 \log^2\frac{1-x}{1+x}\,\frac{dx}x = \int_1^0 \log^2 t \, \frac{-2\,dt}{\frac{1-t}{1+t} (1+t)^2} = 2 \int_0^1 \log^2t \, \frac{dt}{1-t^2} = \cdots$$

The Proof section contains what you want. It starts with the Mittag-Leffler expansion of $\dfrac{\tanh x}x$; replace both instances in the integrand with the series expansion, then evaluate the double sum with Fubini's theorem, converting part of the summand into a simple definite integral over $(0,1)$ along the way.