The value of $$\int_{e^{2}}^{e^{4}} \frac{1}{x}\left(\frac{e^{((log_{e}x)^{2}+1)^{-1}}}{e^{((log_{e}x)^{2}+1)^{-1}}+e^{((6-\log_{e}x)^{2}+1)^{-1}}}\right)dx$$ is

Consider the integral: $$ I = \int_{e^{2}}^{e^{4}} \frac{1}{x} \left( \frac{e^{\left( (\log_{e}x)^{2} + 1 \right)^{-1}}}{e^{\left( (\log_{e}x)^{2} + 1 \right)^{-1}} + e^{\left( (6 - \log_{e}x)^{2} + 1 \right)^{-1}}} \right) dx $$

Substitute $$ t = \log_e x $$, so $$ dt = \frac{1}{x} dx $$.
When $$ x = e^2 $$, $$ t = 2 $$.
When $$ x = e^4 $$, $$ t = 4 $$.
The integral becomes:
$$ I = \int_{2}^{4} \frac{e^{\left( t^{2} + 1 \right)^{-1}}}{e^{\left( t^{2} + 1 \right)^{-1}} + e^{\left( (6 - t)^{2} + 1 \right)^{-1}}} dt $$

Define the function:
$$ f(t) = \frac{e^{\left( t^{2} + 1 \right)^{-1}}}{e^{\left( t^{2} + 1 \right)^{-1}} + e^{\left( (6 - t)^{2} + 1 \right)^{-1}}} $$

Now, compute $$ f(6-t) $$:
$$ f(6-t) = \frac{e^{\left( (6-t)^{2} + 1 \right)^{-1}}}{e^{\left( (6-t)^{2} + 1 \right)^{-1}} + e^{\left( t^{2} + 1 \right)^{-1}}} $$

Adding $$ f(t) $$ and $$ f(6-t) $$:
$$ f(t) + f(6-t) = \frac{e^{a}}{e^{a} + e^{b}} + \frac{e^{b}}{e^{b} + e^{a}} $$
where $$ a = \left( t^{2} + 1 \right)^{-1} $$ and $$ b = \left( (6-t)^{2} + 1 \right)^{-1} $$.
This simplifies to:
$$ f(t) + f(6-t) = \frac{e^{a} + e^{b}}{e^{a} + e^{b}} = 1 $$

Thus, $$ f(t) + f(6-t) = 1 $$.

The integral is:
$$ I = \int_{2}^{4} f(t) dt $$

Consider $$ \int_{2}^{4} f(6-t) dt $$. Substitute $$ u = 6 - t $$, so $$ du = -dt $$.
When $$ t = 2 $$, $$ u = 4 $$; when $$ t = 4 $$, $$ u = 2 $$.
Thus,
$$ \int_{2}^{4} f(6-t) dt = \int_{4}^{2} f(u) (-du) = \int_{2}^{4} f(u) du = I $$

Now, integrate the sum:
$$ \int_{2}^{4} \left[ f(t) + f(6-t) \right] dt = \int_{2}^{4} 1 dt $$
The left side is:
$$ \int_{2}^{4} f(t) dt + \int_{2}^{4} f(6-t) dt = I + I = 2I $$
The right side is:
$$ \int_{2}^{4} 1 dt = t \Big|_{2}^{4} = 4 - 2 = 2 $$

Therefore,
$$ 2I = 2 \implies I = 1 $$

The value of the integral is 1.

Comparing with the options:
A. 2
B. $$ \log_{e}2 $$
C. 1
D. $$ e^{2} $$
The correct answer is option C.