For $x>2,{\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{\sqrt[\phantom{2}]{x^2-4}}{x}dx=?$

$($a$)-\sqrt[\phantom{2}]{x^2-4}+se{c}^{-1}(\frac{x}{2})+C$

$($b$)-x\sqrt[\phantom{2}]{x^2-4}+2se{c}^{-1}(\frac{x}{2})+C$

$($c$)\sqrt[\phantom{2}]{x^2-4}-2se{c}^{-1}(\frac{x}{2})+C$

$($d$)x\sqrt[\phantom{2}]{x^2-4}-se{c}^{-1}(\frac{x}{2})+C$

$($e$)\sqrt[\phantom{2}]{x^2-4}+2se{c}^{-1}(\frac{x}{2})+C$