Let $I =∫\frac{e^x}{e^{4x}+e^{2x}+1}dx$, $J =∫\frac{e^{-x}}{e^{-4x}+e^{-2x}+1}dx$. Then, for an arbitrary constant C, the value of J - I equals

$\frac{1}{2}\log(\frac{e^{2x}-e^{x}+1}{e^{2x}+e^{x}+1})+C$

Here $J=∫\frac{e^{3x}}{e^{4x}+e^{2x}+1}dx$ so that $J-I=∫\frac{e^x(e^{2x}-1)}{e^{4x}+e^{2x}+1}dx$

$=∫\frac{(z^2-1)}{z^4+z^2+1}dz$   where z = e^x

$=∫\frac{(1-\frac{1}{z^2})dz}{(z+\frac{1}{z^2})^2-1}=∫\frac{dt}{t^2-1}$, where  t = z + 1/z.

Hence $J-I\frac{1}{2}\log\frac{t-1}{t+1}+c=\frac{1}{2}\log\frac{e^x+e^{-x}-1}{e^x+e^{-x}+1}+c=\frac{1}{2}ln(\frac{e^x+e^{-x}-1}{e^x+e^{-x}+1})+c$

$J-I=\frac{1}{2}ln(\frac{e^{2x}-e^{x}+1}{e^{2x}+e^{x}+1})+C$

Hence (C) is the correct answer