If $\mathrm{I}_1=\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x, \mathrm{I}_2=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x$, then $\mathrm{I}_2-\mathrm{I}_1=$

• A

  $\frac{1}{2} \log \left(\frac{e^{2 x}-e^{-2 x}+1}{e^{2 x}+e^{-2 x}-1}\right)+\mathrm{c}$
• B

  $\frac{1}{2} \log \left(\frac{e^{2 x}-e^{-2 x}-1}{e^{2 x}+e^{-2 x}+1}\right)+\mathrm{c}$
• C

  $\frac{1}{2} \log \left(\frac{e^{2 x}+e^{-x}+1}{e^{2 x}+e^{-x}-1}\right)+\mathrm{c}$
• Correct answer
  D

  $\frac{1}{2} \log \left(\frac{e^x+e^{-x}-1}{e^x+e^{-x}+1}\right)+\mathrm{c}$

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