Integral of $$$x^{3} - 3$$$

Find $$$\int \left(x^{3} - 3\right)\, dx$$$.

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=3$$$:

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

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

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

Therefore,

$$\int{\left(x^{3} - 3\right)d x} = \frac{x^{4}}{4} - 3 x$$

Simplify:

$$\int{\left(x^{3} - 3\right)d x} = \frac{x \left(x^{3} - 12\right)}{4}$$

Add the constant of integration:

$$\int{\left(x^{3} - 3\right)d x} = \frac{x \left(x^{3} - 12\right)}{4}+C$$

$$$\int \left(x^{3} - 3\right)\, dx = \frac{x \left(x^{3} - 12\right)}{4} + C$$$A