Evaluate the following integral using trigonometric substitution.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{\sqrt[\phantom{2}]{36x^2-49}}{x^3}dx,x>\frac{7}{6}$

What substitution will be the most helpful for evaluating this integral?

A$.x=\frac{7}{6}tan$

B$.x=\frac{7}{6}sec$

C$.x=\frac{7}{6}sin$

Rewrite the given integral using this substitution.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{\sqrt[\phantom{2}]{36x^2-49}}{x^3}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}d$

$$

$($Type an exact answer.$)$