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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 :

JEE · Math · previous-year question

  1. A.loge 4
  2. B.loge 16correct
  3. C.2loge 16
  4. D.4loge (3 + 2$${\sqrt 2 }$$)

Answer

B. loge 16

Explanation

$$\sqrt {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} $$ $$ = \sqrt {{{\left( {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right)}^2}} $$ $$ = \left| {{{x + 1} \over {x - 1}} - {{x - 1} \over {x + 1}}} \right|$$ $$ = \left| {{{{{(x + 1)}^2} - {{(x - 1)}^2}} \over {{x^2} - 1}}} \right|$$ $$ = \left| {{{{x^2} + 2x + 1 - {x^2} + 2x - 1} \over {{x^2} - 1}}} \right|$$ $$ = \left| {{{4x} \over {{x^2} - 1}}} \right|$$ $$\therefore$$ $$I = \int\limits_{ - {1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {{{\left[ {{{\left( {{{x + 1} \over {x - 1}}} \right)}^2} + {{\left( {{{x - 1} \over {x + 1}}} \right)}^2} - 2} \right]}^{{1 \over 2}}}dx} $$ $$I = \int\limits_{{1 \over {\sqrt 2 }}}^{{1 \over {\sqrt 2 }}} {\left| {{{4x} \over {{x^2} - 1}}} \right|dx} $$ 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 $$ \Rightarrow {1 \over {{x^2} - 1}} $$ \Rightarrow {{4x} \over {{x^2} - 1}} $$\therefore$$ $$\left| {{{4x} \over {{x^2} - 1}}} \right| = - {{4x} \over {{x^2} - 1}}$$ when $$x \in \left[ {0,{1 \over {\sqrt 2 }}} \right]$$ $$\therefore$$ $$I = 2\int_0^{{1 \over {\sqrt 2 }}} { - \left( {{{4x} \over {{x^2} - 1}}} \right)dx} $$ $$ = - 4\int_0^{{1 \over {\sqrt 2 }}} {{{2x} \over {{x^2} - 1}}dx} $$ $$ = - 4\left[ {{{\log }_e}\left| {{x^2} - 1} \right|} \right]_0^{{1 \over {\sqrt 2 }}}$$ $$ = - 4\left[ {\log \left| {{1 \over 2} - 1} \right| - \log \left| { - 1} \right|} \right]$$ $$ = - 4{\log _e}\left| { - {1 \over 2}} \right|$$ $$ = - 4{\log _e}{1 \over 2}$$ $$ = - 4\log _e^{{2^{ - 1}}}$$ $$ = 4\log _e^2$$ $$ = \log _e^{{2^4}}$$ $$ = \log _e^{16}$$

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