Question:

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

Updated On: Nov 16, 2024
  • $\frac{\pi}{2} - \log_e 2$
  • $\frac{\pi}{2} + \log_e 2$
  • $\frac{\pi - 1 - \log_e 2}{2}$
  • $\frac{\pi - 1 + \log_e 2}{2}$
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The Correct Option is A

Solution and Explanation

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:

\( \frac{\pi}{2} - \log_e 2 \).

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