Determine.

${\displaystyle \underset{\phantom{}}{\overset{\phantom{}}{}}}(\sqrt[\phantom{2}]{x}-\frac{1}{3x^2})dx$

A$.\frac{3}{2}{x}^{\frac{3}{2}}-\frac{1}{3x}+C$

B$.\frac{2}{3}{x}^{\frac{3}{2}}+\frac{1}{3x}+C$

C$.\frac{1}{2}{x}^{\frac{1}{2}}-\frac{1}{6x^2}+C$

D$.\frac{2}{3}{x}^{\frac{2}{3}}+\frac{1}{6x}+C$

Option D

Option A

Option B

Option C