Step-by-step Solution:

1. Analyze the given integrals We are given two integrals: $I = \int {{{{e^x}} \over {{e^{4x}} + {e^{2x}} + 1}}dx}$ $J = \int {{{{e^{ - x}}} \over {{e^{ - 4x}} + {e^{ - 2x}} + 1}}dx}$ We need to find the value of $J - I$.

2. Simplify the integral J Let's simplify the integrand of J. We can multiply the numerator and the denominator by $e^{4x}$: {{e^{ - x}}} \over {{e^{ - 4x}} + {e^{ - 2x}} + 1}} = {{{e^{ - x}} \cdot {e^{4x}}} \over {({e^{ - 4x}} + {e^{ - 2x}} + 1) \cdot {e^{4x}}}} = {{{e^{3x}}} \over {{e^0} + {e^{2x}} + {e^{4x}}}} = {{{e^{3x}}} \over {{e^{4x}} + {e^{2x}} + 1}} So, the integral J can be written as: $J = \int {{{{e^{3x}}} \over {{e^{4x}} + {e^{2x}} + 1}}dx}$

3. Calculate J - I Now, we can compute $J - I$ by subtracting the integrands: $J - I = \int {{{{e^{3x}}} \over {{e^{4x}} + {e^{2x}} + 1}}dx} - \int {{{{e^x}} \over {{e^{4x}} + {e^{2x}} + 1}}dx}$ J - I = \int {{{e^{3x}} - {e^x}} \over {{e^{4x}} + {e^{2x}} + 1}}dx} We can factor out $e^x$ from the numerator: J - I = \int {{{e^x}({e^{2x}} - 1)} \over {{e^{4x}} + {e^{2x}} + 1}}dx}

4. Perform the integration using substitution Let's use the substitution $u = e^x$. Then, $du = e^x dx$. Substituting these into the integral gives: $J - I = \int {{{u^2 - 1} \over {u^4 + u^2 + 1}}du}$ This is a standard form of integral. We can solve it by dividing the numerator and the denominator by $u^2$: $J - I = \int {{{1 - {1 \over {{u^2}}}} \over {u^2 + 1 + {1 \over {{u^2}}}}}du}$ Now, we manipulate the denominator to relate it to the derivative of a new variable. The numerator is $1 - 1/u^2$, which is the derivative of $u + 1/u$. So we express the denominator in terms of $u + 1/u$: $u^2 + 1 + {1 \over u^2} = \left(u^2 + {1 \over u^2}\right) + 1 = \left(\left(u + {1 \over u}\right)^2 - 2\right) + 1 = \left(u + {1 \over u}\right)^2 - 1$ The integral becomes: $J - I = \int {{{1 - {1 \over {{u^2}}}} \over {{{\left( {u + {1 \over u}} \right)}^2} - 1}}du}$

5. Use a second substitution Let $v = u + {1 \over u}$. Then, $dv = (1 - {1 \over u^2}) du$. The integral simplifies to: $J - I = \int {{{dv} \over {{v^2} - 1}}}$ This is a standard integral formula: $\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \ln \left| \frac{x-a}{x+a} \right| + C$. Here, $a=1$. $J - I = {1 \over 2}\ln \left| {{{v - 1} \over {v + 1}}} \right| + C$

6. Substitute back to the original variable First, substitute back $v = u + {1 \over u}$: $J - I = {1 \over 2}\ln \left| {{{u + {1 \over u} - 1} \over {u + {1 \over u} + 1}}} \right| + C = {1 \over 2}\ln \left| {{{{{u^2} - u + 1} \over u}} \over {{{{u^2} + u + 1} \over u}}} \right| + C = {1 \over 2}\ln \left| {{{{u^2} - u + 1} \over {{u^2} + u + 1}}} \right| + C$ Now, substitute back $u = e^x$: $J - I = {1 \over 2}\ln \left| {{{{({e^x})^2} - {e^x} + 1} \over {{({e^x})^2} + {e^x} + 1}}} \right| + C = {1 \over 2}\ln \left( {{{{e^{2x}} - {e^x} + 1} \over {{e^{2x}} + {e^x} + 1}}} \right) + C$ (Note: The expressions $e^{2x} - e^x + 1 = (e^x - 1/2)^2 + 3/4$ and $e^{2x} + e^x + 1$ are always positive, so the absolute value can be removed.)

7. Compare with the options The calculated value of $J - I$ is ${1 \over 2}\log \left( {{{{e^{2x}} - {e^x} + 1} \over {{e^{2x}} + {e^x} + 1}}} \right) + C$. This matches option C.

Final Answer is option C.