Answer
$y = \frac{2}{3}arcsec(\frac{|2x|}{3})+c$
Work Step by Step
$y' = \frac{1}{x\sqrt{4x^2-9}}$
We can use the basic integration rule $\int\frac{du}{u\sqrt{u^2-a^2}}=\frac{1}{a}arcsec(\frac{|u|}{a})+c$, with $u = 2x$ and $a = 3$.
First, integrating both sides with respect to x, we get $y = \int\frac{1}{x\sqrt{4x^2-9}}$
Factoring out a 2 from the integral, we get $y = 2\int \frac{1}{2x\sqrt{4x^2-9}}$
Now, our integral is in the correct form to apply the rule and we get $y = 2(\frac{1}{3}arcsec(\frac{|2x|}{3}))$.
Finally, simplify to $y = \frac{2}{3}arcsec(\frac{|2x|}{3})+c$
