Evaluate the integral.

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

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

$\frac{(4x^2-4)\sqrt[\phantom{2}]{4x^2-4}}{6}+C$

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

$\frac{(4x^2-4)\sqrt[\phantom{2}]{4x^2-4}}{12}+C$