A) \[\left( \frac{x}{2}-1 \right)\sqrt{1-{{x}^{2}}}+\frac{1}{2}{{\sin }^{-1}}x+C\]
B) \[\left( \frac{x}{2}-1 \right)\sqrt{1-{{x}^{2}}}-\frac{1}{2}{{\sin }^{-1}}x+C\]
C) \[\sqrt{1-{{x}^{2}}}+\frac{1}{2}{{\sin }^{-1}}x+C\]
D) None of the above
Correct Answer: B
Solution :
Let \[I=\int_{{}}^{{}}{x\sqrt{\frac{1-x}{1+x}}dx}\] \[I=\int_{{}}^{{}}{\frac{x}{\sqrt{1-{{x}^{2}}}}}dx+\int_{{}}^{{}}{\frac{-{{x}^{2}}}{\sqrt{1-{{x}^{2}}}}dx}\] \[=-\frac{1}{2}\int_{{}}^{{}}{\frac{-2x}{\sqrt{1-{{x}^{2}}}}}dx+\int_{{}}^{{}}{\frac{1-{{x}^{2}}-1}{\sqrt{1-{{x}^{2}}}}dx}\] \[=-\frac{1}{2}.\frac{{{(1-{{x}^{2}})}^{1/2}}}{1/2}+\int_{{}}^{{}}{\sqrt{1-{{x}^{2}}}}dx-\int_{{}}^{{}}{\frac{dx}{\sqrt{1-{{x}^{2}}}}}\] \[=-\sqrt{1-{{x}^{2}}}+\left\{ \frac{x}{2}\sqrt{1-{{x}^{2}}}+\frac{1}{2}{{\sin }^{-1}}(x) \right\}\] \[-{{\sin }^{-1}}x+C\] \[=(\frac{x}{2}-1)\sqrt{1-{{x}^{2}}}-\frac{1}{2}{{\sin }^{-1}}x+C.\]
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