The value of the integral ∫ - 1 1 log e 1 - x + 1 + x d x is equal to:

$I={\int }_{-1}^1{\log }_e\left(\sqrt{1-x}+\sqrt{1+x}\right)dx$

$\Rightarrow I=2{\int }_0^1\underset{I}{\underbrace{{\log }_e\left(\sqrt{1-x}+\sqrt{1+x}\right)}}·\underset{II}{\underbrace{1}}dx$

$\Rightarrow I=2\left[{\left(x{\log }_e\left(\sqrt{1-x}+\sqrt{1+x}\right)\right)}_0^1-{\int }_0^{1}x·\left(\frac{1}{\sqrt{1-x}+\sqrt{1+x}}\right)·\left(\frac{1}{2\sqrt{1+x}}-\frac{1}{2\sqrt{1-x}}\right)dx\right]$

$\Rightarrow I=2{\log }_e\sqrt{2}-\frac{2}{2}{\int }_0^1\left\{\frac{x\left(\sqrt{1-x}-\sqrt{1+x}\right)}{\left(\sqrt{1-x}+\sqrt{1+x}\right)\sqrt{1-x^2}}\right\}dx$

$\Rightarrow I=2{\log }_e\sqrt{2}-{\int }_0^1\left\{\frac{x{\left(\sqrt{1-x}-\sqrt{1+x}\right)}^2}{-2x\sqrt{1-x^2}}\right\}dx$

$\Rightarrow I=\left({\log }_{e}2\right)+\frac{1}{2}{\int }_0^1\frac{\left(2-2\sqrt{1-x^2}\right)}{\sqrt{1-x^2}} dx$

$\Rightarrow I=\left({\log }_{e}2\right)+{\int }_0^1\left(\frac{1-\sqrt{1-x^2}}{\sqrt{1-x^2}}\right)dx$

$\Rightarrow I=\left({\log }_{e}2\right)+{\left({\sin }^{-1}x-x\right)}_0^1$

$\Rightarrow I=\left({\log }_{\mathrm{e}}2\right)+\frac{\pi }{2}-1$