Given,

$$I = \int {{{{e^x}} \over {{e^{4x}} + {e^{2x}} + 1}}dx} $$

$$J = \int {{{{e^{ - x}}} \over {{e^{ - 4x}} + {e^{ - 2x}} + 1}}dx = \int {{{{e^{3x}}} \over {{e^{4x}} + {e^{2x}} + 1}}dx} } $$

$$\therefore$$ $$J - I = \int {{{{e^x}({e^{2x}} - 1)} \over {{e^{4x}} + {e^{2x}} + 1}}dx} $$

Let $${e^x} = t \Rightarrow {e^x}dx = dt$$

$$\therefore$$ $$J - I = \int {{{{t^2} - 1} \over {{t^4} + {t^2} + 1}}dt} $$

$$ = \int {{{1 - {1 \over {{t^2}}}} \over {{t^2} + 1 + {1 \over {{t^2}}}}}dt} $$

Let, $$t + {1 \over t} = u \Rightarrow \left( {1 - {1 \over {{t^2}}}} \right)dt = du$$

$$J - I = \int {{{du} \over {{u^2} - 1}} = {1 \over 2}\log \left| {{{u - 1} \over {u + 1}}} \right| + C} $$

$$ = {1 \over 2}\log \left| {{{{{{t^2} + 1} \over t} - 1} \over {{{{t^2} + 1} \over t} + 1}}} \right| + C$$

$$ = {1 \over 2}\log \left| {{{{e^{2x}} - {e^x} + 1} \over {{e^{2x}} + {e^x} + 1}}} \right| + C$$