The value of ${\int }_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}{\left({\left(\frac{x+1}{x-1}\right)}^2+{\left(\frac{x-1}{x+1}\right)}^2-2\right)}^{\frac{1}{2}} dx$ is:

We have,

${\int }_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}{\left({\left(\frac{x+1}{x-1}\right)}^2+{\left(\frac{x-1}{x+1}\right)}^2-2\right)}^{\frac{1}{2}} dx$

$={\int }_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\sqrt{{\left[\frac{x-1}{x+1}-\frac{x+1}{x-1}\right]}^2}dx$

$={\int }_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\sqrt{{\left(\frac{-4x}{x^2-1}\right)}^2}dx$

$={\int }_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\sqrt{{\left(\frac{4x}{1-x^2}\right)}^2}dx$

$={\int }_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\sqrt{\frac{16x^2}{{\left(1-x^2\right)}^2}}dx$

$={\int }_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}\left\{\frac{4\left|x\right|}{\left(1-x^2\right)}\right\}dx$

$=2{\int }_0^{\frac{1}{\sqrt{2}}}\left\{\frac{4\left|x\right|}{\left(1-x^2\right)}\right\}dx$

$=4{\int }_0^{\frac{1}{\sqrt{2}}}\left(\frac{2x}{1-x^2}\right)dx$

$=-4{\int }_0^{\frac{1}{\sqrt{2}}}\left(\frac{-2x}{1-x^2}\right)dx$

$=-4{\left[{\log }_e\left(1-x^2\right)\right]}_0^{\frac{1}{\sqrt{2}}}$

$=-4{\log }_e\left(1-\frac{1}{2}\right)$

$=4{\log }_{e}2$

$={\log }_{e}16$