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