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