We want to evaluate
[tex]\displaystyle \int_{-\frac1{\sqrt2}}^{\frac1{\sqrt2}} \sqrt{\left(\frac{x-1}{x+1}\right)^2 + \left(\frac{x+1}{x-1}\right)^2 - 2} \, dx[/tex]
First we note that the integrand is even (replacing x with -x doesn't fundamentally alter the function being integrated), so this is equal to
[tex]\displaystyle 2 \int_0^{\frac1{\sqrt2}} \sqrt{\left(\frac{x-1}{x+1}\right)^2 + \left(\frac{x+1}{x-1}\right)^2 - 2} \, dx[/tex]
The radicand reduces significantly to
[tex]\displaystyle \left(\frac{x-1}{x+1}\right)^2 + \left(\frac{x+1}{x-1}\right)^2 - 2 = \frac{16x^2}{(1-x^2)^2}[/tex]
so that taking the square root, we simplify the integral to
[tex]\displaystyle 8 \int_0^{\frac1{\sqrt2}} \frac x{1-x^2} \, dx[/tex]
which is trivially computed with a substitution of [tex]y = 1 - x^2[/tex] and [tex]dy=-2x\,dx[/tex]:
[tex]\displaystyle -4 \int_1^{\frac12} \frac{dy}y = 4 \int_{\frac12}^1 \frac{dy}y = -4 \ln\left(\frac12\right) = \ln(2^4) = \boxed{\ln(16)}[/tex]
