Integral of $$$\sqrt{3} \sqrt{y}$$$

Find $$$\int \sqrt{3} \sqrt{y}\, dy$$$.

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

$${\color{red}{\int{\sqrt{3} \sqrt{y} d y}}} = {\color{red}{\sqrt{3} \int{\sqrt{y} d y}}}$$

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

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

Therefore,

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

Add the constant of integration:

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

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