\int x {\Tan}^{-1} \sqrt{\frac{1+x^2}{1-x^2}} \dx=

Correct Answer is : x24(π−Cos−1x2)+14√1−x4+c

\frac{x^2}{4} (\pi-{\Cos}^{-1} x^2) +\frac{1}{4} \sqrt{1-x^2}+c

\frac{x^2}{4} (\pi-{\Cos}^{-1} x^2) +\frac{1}{4} \sqrt{1-x^4}+c

\frac{x^2}{4} (\pi+{\Cos}^{-1} x^2) -\frac{1}{4} \sqrt{1-x^4}+c

\frac{x^2}{4} (\pi+{\Cos}^{-1} x^2) -\frac{1}{4} \sqrt{1-x^2}+c

TG EAMCET 3 May 2025 Shift 2 Paper

Question : 74 of 160

Marks: +1, -0

∫xTan−1√‌

1+x2

1−x2

‌dx=

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