When we use trig substitution to evaluate

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{3dx}{x^2\sqrt[\phantom{2}]{x^2+4}}$

which substitution statement do we use?

$x=2*sin,\frac{dx}{d}=2*cos$

$x=2*tan,\frac{dx}{d}=2*se{c}^2$

$x=2*sec,\frac{dx}{d}=2*sec*tan$

$u=x^2+4,\frac{du}{dx}=2x$