We will compute the primitive of the integrand, from which $\rm I$ will fall naturally; some simple algebraic manipulations reveal its true form: $$ \int \frac{\left(\frac{1+x}{1-x}\right)^{1+x}}{\left(\frac{1-x}{1+x}\right)^{1-x}} \diff x = \int \left(\frac{1+x}{1-x}\right)^{1+x} \left(\frac{1+x}{1-x}\right)^{1-x} \diff x = \int \left(\frac{1+x}{1-x}\right)^2 \diff x. $$ This is a simple rational integral with quadratic denominator, so we do not have to leave our confort zone. It thus suffices to divide the numerator by the denominator and then solve a logarithm/arctangent integral: $$ \int \left(\frac{1+x}{1-x}\right)^2 \diff x = \int \left(1 + \frac{4x}{(x-1)^2}\right) \diff x = x + \int \frac{4(x-1) + 4}{(x-1)^2} \diff x = {\color{green}\boxed{x + 4\log|x-1| - \frac{4}{x-1} + C.}} $$
All together, $$ {\rm I} = \left(x + 4\log|x-1| - \frac{4}{x-1}\right)\bigg|_{-1}^0 = (0 + 4\log 1 + 4) - (-1 + 4\log 2 + 2) = {\color{green}\boxed{3 - 4\log 2.}} $$