Integral de $$$\frac{\sqrt{6} \sqrt{x}}{6}$$$

Encontre $$$\int \frac{\sqrt{6} \sqrt{x}}{6}\, dx$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{\sqrt{6}}{6}$$$ e $$$f{\left(x \right)} = \sqrt{x}$$$:

$${\color{red}{\int{\frac{\sqrt{6} \sqrt{x}}{6} d x}}} = {\color{red}{\left(\frac{\sqrt{6} \int{\sqrt{x} d x}}{6}\right)}}$$

Aplique a regra da potência $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=\frac{1}{2}$$$:

$$\frac{\sqrt{6} {\color{red}{\int{\sqrt{x} d x}}}}{6}=\frac{\sqrt{6} {\color{red}{\int{x^{\frac{1}{2}} d x}}}}{6}=\frac{\sqrt{6} {\color{red}{\frac{x^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}}{6}=\frac{\sqrt{6} {\color{red}{\left(\frac{2 x^{\frac{3}{2}}}{3}\right)}}}{6}$$

Portanto,

$$\int{\frac{\sqrt{6} \sqrt{x}}{6} d x} = \frac{\sqrt{6} x^{\frac{3}{2}}}{9}$$

Adicione a constante de integração:

$$\int{\frac{\sqrt{6} \sqrt{x}}{6} d x} = \frac{\sqrt{6} x^{\frac{3}{2}}}{9}+C$$

$$$\int \frac{\sqrt{6} \sqrt{x}}{6}\, dx = \frac{\sqrt{6} x^{\frac{3}{2}}}{9} + C$$$A