question_answer

The value of \[\int_{{}}^{{}}{\frac{dx}{(1+{{x}^{2}})\sqrt{1-{{x}^{2}}}}}\] is *-correct-answer-description-* Let \[\int_{{}}^{{}}{\frac{dx}{(1+{{x}^{2}})\sqrt{1-{{x}^{2}}}}}\] Put \[x=\frac{1}{t}\Rightarrow dx=-\frac{1}{{{t}^{2}}}dt\] Then, \[l=\int_{{}}^{{}}{\frac{-dt}{{{t}^{2}}\left( 1+\frac{1}{{{t}^{2}}} \right)\sqrt{1-{{\left( \frac{1}{t} \right)}^{2}}}}}\] \[=-\int_{{}}^{{}}{\frac{tdt}{({{t}^{2}}+1)\sqrt{{{t}^{2}}-1}}}\]                 Again put \[{{\text{t}}^{\text{2}}}\text{-1 = }{{\text{z}}^{\text{2 }}}\Rightarrow \text{t dt = z dz}\]                 Then  , \[\text{l = -}\int_{{}}^{{}}{\frac{\text{z dz}}{\text{(}{{\text{z}}^{\text{2}}}\text{+2) z}}}\]                 \[=-\int_{{}}^{{}}{\frac{1}{{{z}^{2}}+2}}dz\]                 \[=-\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{z}{\sqrt{2}} \right)+C\]                 \[=-\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{\sqrt{{{t}^{2}}-1}}{\sqrt{2}} \right)+C\]                 \[=-\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{\sqrt{1-{{x}^{2}}}}{\sqrt{2}x} \right)+C\]

A) \[-\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{\sqrt{1-{{x}^{2}}}}{\sqrt{2x}} \right)+C\]

B) \[\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{\sqrt{1-{{x}^{2}}}}{\sqrt{2x}} \right)+C\]

C) \[-{{\tan }^{-1}}\left( \frac{\sqrt{1-{{x}^{2}}}}{\sqrt{2x}} \right)+C\]

D) None of the above