Q. $\int_{- 1}^{0}\frac{dx}{x^{2} + 2x + 2}$ is equal to
A) 0
B) $\pi/4$
C) $\pi/2$
D) $- \pi/4$
Explanation
(b)
$\int_{- 1}^{0}{\frac{dx}{x^{2} + 2x + 2} = \int_{- 1}^{0}\frac{dx}{(x + 1)^{2} + 1}}$
$= \lbrack\tan^{- 1}{(x + 1)\rbrack}_{- 1}^{0}$
$= \lbrack\tan^{- 1}{1 - \tan^{- 1}{0\rbrack = \frac{\pi}{4} - 0 = \frac{\pi}{4}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.