Difference of Indefinite Integrals Involving Exponential Functions

To find the value of $J - I$, we first rewrite the integral $J$ by multiplying the numerator and the denominator by $e^{4x}$:

$J = \int \frac{e^{-x}}{e^{-4x} + e^{-2x} + 1} \, dx = \int \frac{e^{-x} \cdot e^{4x}}{\left(e^{-4x} + e^{-2x} + 1\right) e^{4x}} \, dx = \int \frac{e^{3x}}{e^{4x} + e^{2x} + 1} \, dx$

Now, we compute the difference $J - I$:

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

Substitute $u = e^x$, so $du = e^x \, dx$:

$J - I = \int \frac{u^2 - 1}{u^4 + u^2 + 1} \, du$

Dividing both the numerator and the denominator by $u^2$:

$J - I = \int \frac{1 - \frac{1}{u^2}}{u^2 + 1 + \frac{1}{u^2}} \, du$

We can rewrite the denominator in terms of $\left(u + \frac{1}{u}\right)$:

$u^2 + 1 + \frac{1}{u^2} = \left(u + \frac{1}{u}\right)^2 - 1$

Thus, the integral becomes:

$J - I = \int \frac{1 - \frac{1}{u^2}}{\left(u + \frac{1}{u}\right)^2 - 1} \, du$

Let $v = u + \frac{1}{u}$, so $dv = \left(1 - \frac{1}{u^2}\right) du$:

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

Substituting $v = u + \frac{1}{u}$ back into the expression:

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

Finally, substituting $u = e^x$:

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

Hence, the correct option is C.