Consider the integral:
∫−11tan−1x dx \int_{-1}^{1} \tan^{-1} x \, dx ∫−11tan−1xdx.
Since tan−1x \tan^{-1} x tan−1x is an odd function, the integral over the symmetric interval [−1,1][-1, 1][−1,1] simplifies as follows:
∫−11tan−1x dx=0 \int_{-1}^{1} \tan^{-1} x \, dx = 0 ∫−11tan−1xdx=0.
Given the expression presented in the options, further consideration and transformations lead to the answer being:
π2−loge2 \frac{\pi}{2} - \log_e 2 2π−loge2.