If $\int \frac{1+\mathrm{x}^2}{1+\mathrm{x}^4} \mathrm{dx}=\frac{1}{\sqrt{2}} \tan ^{-1}\left[\frac{\mathrm{f}(\mathrm{x})}{\sqrt{2}}\right]+\mathrm{c}$, then $\mathrm{f}(\mathrm{x})=$

• A

  $x+\frac{1}{x^2}$
• B

  $x-\frac{1}{x^2}$
• C

  $x+\frac{2}{x}$
• Correct answer
  D

  $x-\frac{2}{x}$

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