Question

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

• A. $\frac{\pi}{2} - \log_e 2$
• B. $\frac{\pi}{2} + \log_e 2$
• C. $\frac{\pi - 1 - \log_e 2}{2}$
• D. $\frac{\pi - 1 + \log_e 2}{2}$

Answer 1

Answer: (A)

$\frac{\pi}{2} - \log_e 2$

Answer 2

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:

$\frac{\pi}{2} - \log_e 2$.