The value of ∫_-1^1≤ft|tan^-1x|dx is:

The correct answer is Option (1) → \(\displaystyle\frac{\pi}{2} - \log_e 2\)

Since \(\tan^{-1}(-x) = -\tan^{-1}x\), the function \(\left|\tan^{-1}x\right|\) is even, and \(\tan^{-1}x \ge 0\) for \(x \ge 0\):

    \( \displaystyle I = 2\int_{0}^{1}\tan^{-1}x\,dx \)

Integrating by parts, with \(\tan^{-1}x\) as the first function and 1 as the second:

    \( \displaystyle \int\tan^{-1}x\,dx = x\tan^{-1}x - \int\frac{x}{1 + x^2}\,dx \)

    \( \displaystyle = x\tan^{-1}x - \frac{1}{2}\log_e\left(1 + x^2\right) + C \)

So

    \( \displaystyle I = 2\left[x\tan^{-1}x - \frac{1}{2}\log_e\left(1 + x^2\right)\right]_{0}^{1} \)

    \( \displaystyle = 2\left(\frac{\pi}{4} - \frac{1}{2}\log_e 2\right) \)

    \( \displaystyle = \frac{\pi}{2} - \log_e 2 \)