$$\sqrt {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} $$

$$ = \sqrt {{{\left( {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right)}^2}} $$

$$ = \left| {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right|$$

$$ = \left| {{{{{(x + 1)}^2} - {{(x - 1)}^2}} \over {{x^2} - 1}}} \right|$$

$$ = \left| {{{{x^2} + 2x + 1 - {x^2} + 2x - 1} \over {{x^2} - 1}}} \right|$$

$$ = \left| {{{4x} \over {{x^2} - 1}}} \right|$$

$$\therefore$$ $$I = \int\limits_{ - {1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {{{\left[ {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} \right]}^{{1 \over 2}}}dx} $$

$$I = \int\limits_{{1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx} $$

Let $$f(x) = \left| {{{4x} \over {{x^2} - 1}}} \right|$$

$$\therefore$$ $$f( - x) = \left| {{{4( - x)} \over {{{( - x)}^2} - 1}}} \right| = \left| {{{ - 4x} \over {{x^2} - 1}}} \right| = \left| {{{4x} \over {{x^2} - 1}}} \right|$$

$$ \Rightarrow f(x) = f( - x)$$

$$\therefore$$ f(x) is a even function.

$$\therefore$$ $$I = 2\int_0^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx} $$

Using property, If f(x) is an even function then,

$$\int_{ - a}^a {f(x) = 2\int_0^a {f(x)dx} } $$

x > 0 when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$

$$\Rightarrow$$ x² > 0 when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$

$$\Rightarrow$$ x² $$-$$ 1 < 0 when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$

$$ \Rightarrow {1 \over {{x^2} - 1}} < 0$$ when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$

$$ \Rightarrow {{4x} \over {{x^2} - 1}} < 0$$ when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$

$$\therefore$$ $$\left| {{{4x} \over {{x^2} - 1}}} \right| = - {{4x} \over {{x^2} - 1}}$$ when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$

$$\therefore$$ $$I = 2\int_0^{{1 \over {\sqrt 2 }}} { - \left( {{{4x} \over {{x^2} - 1}}} \right)dx} $$

$$ = - 4\int_0^{{1 \over {\sqrt 2 }}} {{{2x} \over {{x^2} - 1}}dx} $$

$$ = - 4\left[ {{{\log }_e}\left| {{x^2} - 1} \right|} \right]_0^{{1 \over {\sqrt 2 }}}$$

$$ = - 4\left[ {\log \left| {{1 \over 2} - 1} \right| - \log \left| { - 1} \right|} \right]$$

$$ = - 4{\log _e}\left| { - {1 \over 2}} \right|$$

$$ = - 4{\log _e}{1 \over 2}$$

$$ = - 4\log _e^{{2^{ - 1}}}$$

$$ = 4\log _e^2$$

$$ = \log _e^{{2^4}}$$

$$ = \log _e^{16}$$