Question $9$

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}\frac{dx}{4x^2+9}=$

$\frac{1}{3}ta{n}^{-1}(\frac{x}{3})+C$

$\frac{1}{6}ta{n}^{-1}(\frac{2x}{3})+C$

$\frac{1}{3}ta{n}^{-1}(\frac{2x}{3})+C$

$\frac{1}{2}ta{n}^{-1}(\frac{2x}{3})+C$