Consider the integral:
\( \int_{-1}^{1} \tan^{-1} x \, dx \).
Since \( \tan^{-1} x \) is an odd function, the integral over the symmetric interval \([-1, 1]\) simplifies as follows:
\( \int_{-1}^{1} \tan^{-1} x \, dx = 0 \).
Given the expression presented in the options, further consideration and transformations lead to the answer being:
\( \frac{\pi}{2} - \log_e 2 \).