$\displaystyle \int \frac{\sqrt {4x^2-16}}{x} dx

$

$\displaystyle = \int \frac {\sqrt{(2x)^2 - 4^2}} {x} dx$

2x = 4 sec t

x = 2 sec t

$\displaystyle \int \frac {\sqrt {4(2sect)^2 - 16}} {2sect} dx$

$\displaystyle \int \frac {\sqrt {16(sec^2t-1}}{2sect} dx$

$\displaystyle \int \frac {4 \sqrt {sec^2t-1}}{2sect} dx$

How can I proceed? I don't think there's an identity to simplify sec^2 t - 1.