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

  • A
    $\ln 2$
  • B
    $2$
  • C
    $1$
  • D
    $e^2$

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