If\[\int{{{\log }_{e}}\left( \sqrt{1-x}+\sqrt{1+x} \right)dx}\]\[=x{{\log }_{e}}\left( \sqrt{1-x}+\sqrt{1+x} \right)+g(x)+C\]. Then \[g(x)=\]

If\[\int{{{\log }_{e}}\left( \sqrt{1-x}+\sqrt{1+x} \right)dx}\]\[=x{{\

1.	If\[\int{{{\log }_{e}}\left( \sqrt{1-x}+\sqrt{1+x} \right)dx}\]\[=x{{\log }_{e}}\left( \sqrt{1-x}+\sqrt{1+x} \right)+g(x)+C\]. Then \[g(x)=\]
A.	\[x-{{\sin }^{-1}}x\]
B.	\[{{\sin }^{-1}}x-x\]
C.	\[x+{{\sin }^{-1}}x\]
D.	\[{{\sin }^{-1}}x-{{x}^{2}}\]
Answer» C. \[x+{{\sin }^{-1}}x\]

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If\[\int{{{\log }_{e}}\left( \sqrt{1-x}+\sqrt{1+x} \right)dx}\]\[=x{{\log }_{e}}\left( \sqrt{1-x}+\sqrt{1+x} \right)+g(x)+C\]. Then \[g(x)=\]

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