Q. $\int_{}^{}\frac{dx}{2\sqrt{x}(1 + x)}$ is equal to
A) $\frac{1}{2}\tan^{- 1}{\left( \sqrt{x} \right) + c}$
B) $\tan^{- 1}{\left( \sqrt{x} \right) + c}$
C) $2\tan^{- 1}{\left( \sqrt{x} \right) + c}$
D) None of these
Explanation
(b)
Let $I = \int_{}^{}\frac{dx}{2\sqrt{x}(1 + x)}$
Put $\sqrt{x} = t\ \Rightarrow \ \frac{1}{2\sqrt{x}}dx = dt$
$\therefore I = \int_{}^{}{\frac{dt}{1 + t^{2}} = \tan^{- 1}{t + c = \tan^{- 1}{\sqrt{x} + c}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.