$\int x \operatorname{Tan}^{-1} \sqrt{\frac{1+x^2}{1-x^2}} d x=$

• A

  $\frac{x^2}{4}\left(\pi-\operatorname{Cos}^{-1} x^2\right)+\frac{1}{4} \sqrt{1-x^2}+\mathrm{c}$
• Correct answer
  B

  $\frac{x^2}{4}\left(\pi-\operatorname{Cos}^{-1} x^2\right)+\frac{1}{4} \sqrt{1-x^4}+c$
• C

  $\frac{x^2}{4}\left(\pi+\operatorname{Cos}^{-1} x^2\right)-\frac{1}{4} \sqrt{1-x^4}+\mathrm{c}$
• D

  $\frac{x^2}{4}\left(\pi+\operatorname{Cos}^{-1} x^2\right)-\frac{1}{4} \sqrt{1-x^2}+\mathrm{c}$

Hence, the correct answer is option: 'B'.