$\quad\quad\quad\quad\quad\quad\quad\quad(I.B.P.)$
$\therefore I=\mid x \cdot I n(\sqrt{1-x}+\sqrt{1-x})_{0}^{1}$
$\left.-\int_{0}^{1} x \cdot\left(\frac{1}{\sqrt{1-x}+\sqrt{1+x}}\right) \cdot\left(\frac{1}{2 \sqrt{1+x}}-\frac{1}{2 \sqrt{1-x}}\right) d x\right]$
$=2(\operatorname{In} \sqrt{2}-0)-\frac{2}{2} \int_{0}^{1} \frac{x \sqrt{1-x}-\sqrt{1+x} d x}{\sqrt{1-x}+\sqrt{1+x} \sqrt{1-x^{2}}}$
$=2\left(\log _{e} 2\right) \int_{0}^{1} \frac{x \cdot\left(2-2 \sqrt{1-x^{2}}\right)}{-2 x \sqrt{1-x^{2}}} d x$
(After rationalisation)
$=\left(\log _{e} 2\right)+\int_{0}^{1}\left(\frac{\left(1-\sqrt{1-x^{2}}\right)}{\sqrt{1-x^{2}}}\right) d x$
$=\left(\log _{e} 2\right)+\left(\sin ^{-1} x\right)_{0}^{1}-1$
$=\log _{e} 2+\left(\frac{\pi}{2}-0\right)-1$
$\therefore I=(\log 2)+\frac{\pi}{2}-1$
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