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