Answer
$\frac{1}{\sqrt 3}\sec^{-1}(2x)+c$
Work Step by Step
Since $u=2x$, then $du= 2dx$. Now, we have
$$\int \frac{ dx}{x\sqrt{12x^2-3}}=\frac{1}{\sqrt 3}\int \frac{ du}{u\sqrt{u^2-1}}\\=\frac{1}{\sqrt 3}\sec^{-1}u+c=\frac{1}{\sqrt 3}\sec^{-1}(2x)+c
.$$
