The value of \(\int_{0}^{1}\displaystyle\tan^{-1}x\,dx\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{\pi}{4}-\frac{1}{2}\log_e 2\)
By integration by parts, taking \(\tan^{-1}x\) as the first function and 1 as the second,
\( \displaystyle \int_{0}^{1}\tan^{-1}x\,dx = \left[x\tan^{-1}x\right]_{0}^{1} - \int_{0}^{1}\frac{x}{1+x^{2}}\,dx \)
The first part is \(1\cdot\frac{\pi}{4} - 0 = \frac{\pi}{4}\). For the second part,
\( \displaystyle \int_{0}^{1}\frac{x}{1+x^{2}}\,dx = \frac{1}{2}\left[\log_e(1+x^{2})\right]_{0}^{1} = \frac{1}{2}\log_e 2 \)
Hence
\( \displaystyle \int_{0}^{1}\tan^{-1}x\,dx = \frac{\pi}{4} - \frac{1}{2}\log_e 2 \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{\pi}{4}+\frac{1}{2}\log_e 2\) — Incorrect. In integration by parts the second integral is subtracted, not added.
Option (3) \(\displaystyle\frac{\pi}{4}-\log_e 2\) — Incorrect. It leaves out the factor \(\frac{1}{2}\) in \(\int\frac{x}{1+x^{2}}\,dx = \frac{1}{2}\log_e(1+x^{2})\).
Option (4) \(\displaystyle\frac{\pi}{4}\) — Incorrect. This is only the first part \(\left[x\tan^{-1}x\right]_{0}^{1}\); the second integral is left out.