$\int\left(\frac{1-\log x}{1+(\log x)^2}\right)^2 d x=$

• A

  $\frac{1}{1+(\log x)^2}+c$
• B

  $\frac{\log x}{1+(\log x)^2}+c$
• Correct answer
  C

  $\frac{x}{1+(\log x)^2}+c$
• D

  $\frac{x^2}{1+(\log x)^2}+\mathrm{c}$

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