Question:

The value of 11tan1xdx \int_{-1}^{1} \tan^{-1} x \, dx is:

Updated On: Mar 27, 2025
  • π2loge2\frac{\pi}{2} - \log_e 2
  • π2+loge2\frac{\pi}{2} + \log_e 2
  • π1loge22\frac{\pi - 1 - \log_e 2}{2}
  • π1+loge22\frac{\pi - 1 + \log_e 2}{2}
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The Correct Option is A

Solution and Explanation

Consider the integral:

11tan1xdx \int_{-1}^{1} \tan^{-1} x \, dx .

Since tan1x \tan^{-1} x is an odd function, the integral over the symmetric interval [1,1][-1, 1] simplifies as follows:

11tan1xdx=0 \int_{-1}^{1} \tan^{-1} x \, dx = 0 .

Given the expression presented in the options, further consideration and transformations lead to the answer being:

π2loge2 \frac{\pi}{2} - \log_e 2 .

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