The value of the integral $\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt {1 - x} + \sqrt {1 + x} )dx}$ is equal to:

$\int\limits_{ - 1}^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} )dx}$<br/><br/>We know, $\int\limits_{ - a}^a {f(x)dx = 2\int\limits_0^a {f(x)dx} }$<br/><br/>So, $2\int\limits_0^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} )dx}$<br/><br/>$l = 2\int\limits_0^1 {{{\log }_e}(\sqrt 1 - x + \sqrt {1 + x} ).1dx}$<br/><br/>$$ \Rightarrow {l \over 2} = {\log _e}(\sqrt {1 - x} + \sqrt {1 + x} ).\,x]_0^1 - \int\limits_0^1 {{{{1 \over {2\sqrt {1 + x} }} - {1 \over {2\sqrt {1 - x} }}} \over {\sqrt {1 - x} + \sqrt {1 + x} }}.\,x\,dx} $$<br/><br/>$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {\left( {{{\sqrt {1 - x} - \sqrt {1 + x} } \over {\sqrt {1 - x} + \sqrt {1 + x} }}} \right){x \over {\sqrt {1 - {x^2}} }}dx} $$<br/><br/>$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {\left( {{{(1 - x) + (1 + x) - 2\sqrt {1 - {x^2}} } \over {(1 - x) - (1 + x)}}} \right){x \over {\sqrt {1 - {x^2}} }}dx} $$<br/><br/>$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 - {1 \over 2}\int\limits_0^1 {{{2(1 - \sqrt {1 - {x^2}} )} \over { - 2x}}.{x \over {\sqrt {1 - {x^2}} }}dx} $$<br/><br/>$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}\int\limits_0^1 {\left( {{1 \over {\sqrt {1 - {x^2}} }} - 1} \right)dx} $$<br/><br/>$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}[{\sin ^{ - 1}}x - x]_0^1$$<br/><br/>$$ \Rightarrow {l \over 2} = {\log _e}\sqrt 2 + {1 \over 2}\left( {{\pi \over 2} - 1} \right)$$<br/><br/>$\therefore$ $l = {\log _e}2 + {\pi \over 2} - 1$