Evaluate the following integral using trigonometric substitution.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{\sqrt[\phantom{2}]{16x^2-25}}{x^3}dx,x>\frac{5}{4}$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{\sqrt[\phantom{2}]{16x^2-25}}{x^3}dx=$

$(Typean$ exact answer.$)$