Evaluate the following integral.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x^2-4x4}{\sqrt[\phantom{2}]{x^2-4x8}}dx$

Rewrite the integrand by completing the square.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{x^2-4x4}{\sqrt[\phantom{2}]{x^2-4x8}}dx={\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}},dx$

$($Simplify your answer.$)$