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

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

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

$\text{Find } J-I$

$\text{1. Rewrite } J \text{ in positive powers}$

$\text{2. Form one integral for } J-I$

$\text{3. Substitute } t=e^{x} \text{ and integrate}$

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

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

$t=e^{x},\quad dt=e^{x}\,dx$

$e^{4x}+e^{2x}+1 = t^{4}+t^{2}+1$

$e^{3x}dx=t^{2}dt,\qquad e^{x}dx=dt$

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

$=\int \dfrac{1-1/t^{2}}{t^{2}+1+1/t^{2}}\,dt$

$v=t+\dfrac{1}{t},\quad dv=\left(1-\dfrac{1}{t^{2}}\right)dt$

$t^{2}+1+\dfrac{1}{t^{2}} = \left(t+\dfrac{1}{t}\right)^{2}-1 = v^{2}-1$

$J-I=\int \dfrac{dv}{v^{2}-1} = \dfrac{1}{2}\log\left|\dfrac{v-1}{v+1}\right|+C$

$v=e^{x}+e^{-x}$

$\dfrac{v-1}{v+1}=\dfrac{e^{2x}-e^{x}+1}{e^{2x}+e^{x}+1}$

$\Rightarrow\ \text{option (3)}$

$\boxed{\text{Answer: option (3)}}$