Q. $\int_{}^{}\frac{dx}{x + \sqrt{x}}$ is equal to
A) $\frac{1}{2}\log{\left( 1 + \sqrt{x} \right) + c}$
B) $2\log{\left( 1 + \sqrt{x} \right) + c}$
C) $\frac{1}{4}\log{\left( 1 + \sqrt{x} \right) + c}$
D) $3\log{\left( 1 + \sqrt{x} \right) + c}$
Explanation
(b)
Let $I = \int_{}^{}\frac{dx}{x + \sqrt{x}} = \int_{}^{}\frac{1}{\sqrt{x}\left( 1 + \sqrt{x} \right)}dx$
Put $1 + \sqrt{x} = t\ \Rightarrow \ \ \frac{1}{\sqrt{x}}dx = 2dt$
$\therefore\ I = 2\int_{}^{}\frac{1}{t}\ dt = 2\log t + c$
$= 2\log{(1 + \sqrt{x})} + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.