Integral of $$$\frac{1}{3 y^{\frac{2}{3}}}$$$

Find $$$\int \frac{1}{3 y^{\frac{2}{3}}}\, dy$$$.

Apply the constant multiple rule $$$\int c f{\left(y \right)}\, dy = c \int f{\left(y \right)}\, dy$$$ with $$$c=\frac{1}{3}$$$ and $$$f{\left(y \right)} = \frac{1}{y^{\frac{2}{3}}}$$$:

$${\color{red}{\int{\frac{1}{3 y^{\frac{2}{3}}} d y}}} = {\color{red}{\left(\frac{\int{\frac{1}{y^{\frac{2}{3}}} d y}}{3}\right)}}$$

Apply the power rule $$$\int y^{n}\, dy = \frac{y^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=- \frac{2}{3}$$$:

$$\frac{{\color{red}{\int{\frac{1}{y^{\frac{2}{3}}} d y}}}}{3}=\frac{{\color{red}{\int{y^{- \frac{2}{3}} d y}}}}{3}=\frac{{\color{red}{\frac{y^{- \frac{2}{3} + 1}}{- \frac{2}{3} + 1}}}}{3}=\frac{{\color{red}{\left(3 y^{\frac{1}{3}}\right)}}}{3}=\frac{{\color{red}{\left(3 \sqrt[3]{y}\right)}}}{3}$$

Therefore,

$$\int{\frac{1}{3 y^{\frac{2}{3}}} d y} = \sqrt[3]{y}$$

Add the constant of integration:

$$\int{\frac{1}{3 y^{\frac{2}{3}}} d y} = \sqrt[3]{y}+C$$

$$$\int \frac{1}{3 y^{\frac{2}{3}}}\, dy = \sqrt[3]{y} + C$$$A