The integral $${{24} \over \pi }\int_0^{\sqrt 2 } {{{(2 - {x^2})dx} \over {(2 + {x^2})\sqrt {4 + {x^4}} }}} $$ is equal to ____________.

<p>$$I = {{24} \over \pi }\int_0^{\sqrt 2 } {{{2 - {x^2}} \over {(2 + {x^2})\sqrt {4 + {x^4}} }}dx} $$</p> <p>Let $x = \sqrt 2 t \Rightarrow dx = \sqrt 2 dt$</p> <p>$$I = {{24} \over \pi }\int_0^1 {{{(2 - 2{t^2})\,.\,\sqrt 2 dt} \over {(2 + 2{t^2})\sqrt {4 + 4{t^4}} }}} $$</p> <p>$$ = {{12\sqrt 2 } \over \pi }\int_0^1 {{{\left( {{1 \over {{t^2}}} - 1} \right)dt} \over {\left( {t + {1 \over t}} \right)\sqrt {{{\left( {t + {1 \over t}} \right)}^2} - 2} }}} $$</p> <p>Let $t + {1 \over t} = u$</p> <p>$\Rightarrow \left( {1 - {1 \over {{t^2}}}} \right)dt = du$</p> <p>$$ = {{12\sqrt 2 } \over \pi }\int_\infty ^2 {{{ - du} \over {u\sqrt {{4^2} - 2} }}} $$</p> <p>$$ = {{12\sqrt 2 } \over \pi }\int_2^\infty {{{du} \over {{u^2}\sqrt { - {{\left( {{{\sqrt 2 } \over u}} \right)}^2}} }}} $$</p> <p>$$ = {{12\sqrt 2 } \over \pi }\int_{{1 \over {\sqrt 2 }}}^0 {{{ - {1 \over {\sqrt 2 }}dp} \over {\sqrt {1 - {p^2}} }}} $$</p> <p>$= {{12} \over \pi }\left[ {{{\sin }^{ - 1}}p} \right]_0^{{1 \over {\sqrt 2 }}}$</p> <p>$= {{12} \over \pi }\,.\,{\pi \over 4} = 3$</p>