$∫\frac{1}{|x|\sqrt{x^2-9}}dx$ is equal to (given that C is constant of integration )
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{1}{3}sec^{-1}\frac{x}{3}+C$
$∫\frac{1}{|x|\sqrt{x^2-9}}dx=\frac{1}{3}∫\frac{dx}{|x|\sqrt{(\frac{x}{3})^2}-1}$
let $y=\frac{x}{3}⇒x=3y⇒\frac{dx}{3}=dy$
so $\frac{1}{3}∫\frac{dy}{|y|\sqrt{y^2-1}}=\frac{1}{3}\sec^{-1}(y)+C$
$=\frac{1}{3}\sec^{-1}(\frac{x}{3})+C$