Answer
$$F(x) =2 x^{3/2}-\frac{3}{2}x^{4/3}+C$$
Work Step by Step
Given $$f(x) = 3\sqrt{x}-2\sqrt[3]{x}= 3x^{1/2}-2x^{1/3}$$
Then by using if $f(x)=x^{\alpha}\ \to\ \ F(x) = \frac{x^{\alpha+1}}{\alpha+1}+C$
\begin{align*}
F(x) &= \frac{ 3}{3/2}x^{3/2}-\frac{2}{4/3}x^{4/3}+C\\
&=2 x^{3/2}-\frac{3}{2}x^{4/3}+C
\end{align*}
To check
\begin{align*}
F'(x) &=3x^{\frac{1}{2}}-2x^{\frac{1}{3}}\\
&= 3\sqrt{x}-2\sqrt[3]{x}\\
&=f(x)
\end{align*}
