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