$$\displaystyle\int{ \frac{ \sqrt{ 16 { x }^{ 2 } -9 } }{ x } }d x$$
$-3\arctan(\frac{\sqrt{16x^{2}-9}}{3})+\sqrt{16x^{2}-9}+С$
$\frac{\sqrt{16x^{2}-9}}{x}$
