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