$\int x \operatorname{Cos}^{-1}\left(\frac{1-x^2}{1+x^2}\right) d x(x>0)=$

• Correct answer
  A

  $-x+\left(1+x^2\right) \operatorname{Tan}^{-1} x+c$
• B

  $x-\left(1+x^2\right) \operatorname{Cot}^{-1} x+c$
• C

  $-x+\left(1+x^2\right) \operatorname{Cot}^{-1} x+c$
• D

  $x-\left(1+x^2\right) \operatorname{Tan}^{-1} x+c$

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