Q. If $\int_{}^{}\frac{dx}{(x + 100)\sqrt{x + 99}} = f(x) + c$, then $f(x)$ is
A) $2(x + 100)^{1/2\ }$
B) $3(x + 100)^{1/2}$
C) $2\tan^{- 1}{(\sqrt{x + 99})}$
D) $2\tan^{- 1}{(\sqrt{x + 100})}$
Explanation
(c)
Let $I = \int_{}^{}\frac{dx}{\left\lbrack \left( \sqrt{x + 99} \right)^{2} + 1 \right\rbrack\sqrt{x + 99}}$
Put $\sqrt{x + 99} = t\ \ \Rightarrow \ \ \frac{1}{\sqrt{x + 99}}dx = 2dt$
$\therefore\ I = \int_{}^{}\frac{2dt}{t^{2} + 1} = 2\tan^{- 1}t + c$
$= 2\tan^{- 1}\sqrt{x + 99} + c$
But $2\tan^{- 1}\sqrt{x + 99} + c = f(x) + c$ [given]
$\therefore\ f(x) = 2\tan^{- 1}\sqrt{x + 99}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.