Q. $\int_{}^{}\frac{dx}{x\sqrt{x^{6} - 16}}$ is equal to
A) $\frac{1}{3}\sec^{- 1}{\left( \frac{x^{3}}{4} \right) + c}$
B) $\cos^{- 1}\left( \frac{x^{3}}{4} \right) + c$
C) $\frac{1}{12}\sec^{- 1}\left( \frac{x^{3}}{4} \right) + c$
D) $\sec^{- 1}\left( \frac{x^{3}}{4} \right) + c$
Explanation
(c)
Let $I = \int_{}^{}\frac{dx}{x\sqrt{x^{6} - 16}}$
$= \frac{1}{3}\int_{}^{}\frac{3x^{2}}{x^{3}\sqrt{\left( x^{3} \right)^{2} - 4^{2}}}dx$
Put $x^{3} = t\ \ \Rightarrow 3x^{2}dx = dt$
$\therefore\ I = \frac{1}{3}\int_{}^{}\frac{dt}{t\sqrt{t^{2} - 4^{2}}}$
$= \frac{1}{3 \times 4}\sec^{- 1}\left( \frac{t}{4} \right) + c$
$= \frac{1}{12}\sec^{- 1}{\left( \frac{x^{3}}{4} \right) + c}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.