Q. $\int_{}^{}\sqrt{\frac{x - 1}{x + 1}\ dx}$ is equal to
A) $2\sqrt{x^{2} + 1} + \sin^{- 1}{x + c}$
B) $\sqrt{x^{2} - 1} - \sin^{- 1}{x + c}$
C) $2\sqrt{x^{2} - 1} + \sin^{- 1}{x + c}$
D) $\sqrt{\frac{x^{2} - 1}{2} + \sin^{- 1}{x + c}}$
Explanation
(b)
Let $I = \int_{}^{}\sqrt{\frac{x - 1}{x + 1}}dx = \int_{}^{}\frac{x - 1}{\sqrt{x^{2} - 1}}\ dx$
$= \int_{}^{}\frac{x}{\sqrt{x^{2} - 1}}dx - \int_{}^{}\frac{1}{\sqrt{x^{2} - 1}}\ dx$
$= \sqrt{x^{2} - 1} - \sin^{- 1}{x + c}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.