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

<p>$$\sqrt {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} $$</p> <p>$$ = \sqrt {{{\left( {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right)}^2}} $$</p> <p>$= \left| {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right|$</p> <p>$= \left| {{{{{(x + 1)}^2} - {{(x - 1)}^2}} \over {{x^2} - 1}}} \right|$</p> <p>$= \left| {{{{x^2} + 2x + 1 - {x^2} + 2x - 1} \over {{x^2} - 1}}} \right|$</p> <p>$= \left| {{{4x} \over {{x^2} - 1}}} \right|$</p> <p>$\therefore$ $$I = \int\limits_{ - {1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {{{\left[ {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} \right]}^{{1 \over 2}}}dx} $$</p> <p>$$I = \int\limits_{{1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx} $$</p> <p>Let $f(x) = \left| {{{4x} \over {{x^2} - 1}}} \right|$</p> <p>$\therefore$ $$f( - x) = \left| {{{4( - x)} \over {{{( - x)}^2} - 1}}} \right| = \left| {{{ - 4x} \over {{x^2} - 1}}} \right| = \left| {{{4x} \over {{x^2} - 1}}} \right|$$</p> <p>$\Rightarrow f(x) = f( - x)$</p> <p>$\therefore$ f(x) is a even function.</p> <p>$\therefore$ $I = 2\int_0^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx}$</p> <p>Using property, If f(x) is an even function then,</p> <p>$\int_{ - a}^a {f(x) = 2\int_0^a {f(x)dx} }$</p> <p>x > 0 when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$</p> <p>$\Rightarrow$ x² > 0 when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$</p> <p>$\Rightarrow$ x² $-$ 1 < 0 when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$</p> <p>$\Rightarrow {1 \over {{x^2} - 1}} < 0$ when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$</p> <p>$\Rightarrow {{4x} \over {{x^2} - 1}} < 0$ when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$</p> <p>$\therefore$ $\left| {{{4x} \over {{x^2} - 1}}} \right| = - {{4x} \over {{x^2} - 1}}$ when $x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$</p> <p>$\therefore$ $$I = 2\int_0^{{1 \over {\sqrt 2 }}} { - \left( {{{4x} \over {{x^2} - 1}}} \right)dx} $$</p> <p>$= - 4\int_0^{{1 \over {\sqrt 2 }}} {{{2x} \over {{x^2} - 1}}dx}$</p> <p>$$ = - 4\left[ {{{\log }_e}\left| {{x^2} - 1} \right|} \right]_0^{{1 \over {\sqrt 2 }}}$$</p> <p>$$ = - 4\left[ {\log \left| {{1 \over 2} - 1} \right| - \log \left| { - 1} \right|} \right]$$</p> <p>$= - 4{\log _e}\left| { - {1 \over 2}} \right|$</p> <p>$= - 4{\log _e}{1 \over 2}$</p> <p>$= - 4\log _e^{{2^{ - 1}}}$</p> <p>$= 4\log _e^2$</p> <p>$= \log _e^{{2^4}}$</p> <p>$= \log _e^{16}$</p>