Integral of $$$\frac{\sqrt{3} x^{8}}{4}$$$

Find $$$\int \frac{\sqrt{3} x^{8}}{4}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{\sqrt{3}}{4}$$$ and $$$f{\left(x \right)} = x^{8}$$$:

$${\color{red}{\int{\frac{\sqrt{3} x^{8}}{4} d x}}} = {\color{red}{\left(\frac{\sqrt{3} \int{x^{8} d x}}{4}\right)}}$$

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=8$$$:

$$\frac{\sqrt{3} {\color{red}{\int{x^{8} d x}}}}{4}=\frac{\sqrt{3} {\color{red}{\frac{x^{1 + 8}}{1 + 8}}}}{4}=\frac{\sqrt{3} {\color{red}{\left(\frac{x^{9}}{9}\right)}}}{4}$$

Therefore,

$$\int{\frac{\sqrt{3} x^{8}}{4} d x} = \frac{\sqrt{3} x^{9}}{36}$$

Add the constant of integration:

$$\int{\frac{\sqrt{3} x^{8}}{4} d x} = \frac{\sqrt{3} x^{9}}{36}+C$$

$$$\int \frac{\sqrt{3} x^{8}}{4}\, dx = \frac{\sqrt{3} x^{9}}{36} + C$$$A