A) \[-{{\sin }^{-1}}x-\sqrt{1-{{x}^{2}}}\,+c\]
B) \[{{\sin }^{-1}}x+\sqrt{1-{{x}^{2}}}\,+c\]
C) \[{{\sin }^{-1}}x-\sqrt{1-{{x}^{2}}}\,+c\]
D) \[-{{\sin }^{-1}}x-\sqrt{{{x}^{2}}-1}\,+c\]
Correct Answer: C
Solution :
\[I=\int{\sqrt{\frac{1+x}{1-x}}dx}\]\[=\int{\frac{1+x}{\sqrt{1-{{x}^{2}}}}dx}\] \[=\int{\frac{dx}{\sqrt{1-{{x}^{2}}}}+\int{\frac{x}{\sqrt{1-{{x}^{2}}}}dx}}\]\[={{\sin }^{-1}}x-\sqrt{1-{{x}^{2}}}+c\].
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