The value of \(\int_{0}^{1}\displaystyle\frac{dx}{x^{2}+2x+2}\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\tan^{-1}2-\displaystyle\frac{\pi}{4}\)
Completing the square, \(x^{2}+2x+2 = (x+1)^{2}+1\). So
\( \displaystyle \int_{0}^{1}\frac{dx}{(x+1)^{2}+1} = \left[\tan^{-1}(x+1)\right]_{0}^{1} \)
\( = \tan^{-1}2 - \tan^{-1}1\)
\( \displaystyle = \tan^{-1}2 - \frac{\pi}{4} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{\pi}{4}\) — Incorrect. This is only \(\tan^{-1}1\), the value at the lower limit.
Option (2) \(\tan^{-1}2\) — Incorrect. It leaves out the value at the lower limit: at \(x = 0\), \(\tan^{-1}(0+1) = \frac{\pi}{4}\), not 0.
Option (3) \(\tan^{-1}3-\displaystyle\frac{\pi}{4}\) — Incorrect. At the upper limit \(x = 1\), \(x+1 = 2\), not 3.