∫$$dx1+x21$$−$$x2is$$ equal to

Correct option is 1−12tan−1⁡1−x22x+C

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$\int \frac{dx}{(1 + x^{2}) \sqrt{1 - x^{2}}}$ is equal to

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−12tan−1⁡1−x22x+C

12tan−1⁡1−x22x+C

−tan−1⁡1−x22x+C

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detailed solution

Correct option is A

Put x=1t⇒dx=−1t2dt∴ I=∫−dtt21+1t21−1t2=−∫tdtt2+1t2−1Again, put t2−1=y2⇒tdt=ydy∴ l=−∫ydyy2+2y=−12tan−1⁡y2+C=−12tan−1⁡1−x22x+C

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∫dx1+x21−x2is equal to

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$\int \frac{dx}{(1 + x^{2}) \sqrt{1 - x^{2}}}$ is equal to