Integral of $$$\frac{3 x^{3}}{16}$$$

Find $$$\int \frac{3 x^{3}}{16}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{3}{16}$$$ and $$$f{\left(x \right)} = x^{3}$$$:

$${\color{red}{\int{\frac{3 x^{3}}{16} d x}}} = {\color{red}{\left(\frac{3 \int{x^{3} d x}}{16}\right)}}$$

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

$$\frac{3 {\color{red}{\int{x^{3} d x}}}}{16}=\frac{3 {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}}{16}=\frac{3 {\color{red}{\left(\frac{x^{4}}{4}\right)}}}{16}$$

Therefore,

$$\int{\frac{3 x^{3}}{16} d x} = \frac{3 x^{4}}{64}$$

Add the constant of integration:

$$\int{\frac{3 x^{3}}{16} d x} = \frac{3 x^{4}}{64}+C$$

$$$\int \frac{3 x^{3}}{16}\, dx = \frac{3 x^{4}}{64} + C$$$A