Integral of $$$2 x^{3} - 9$$$

Find $$$\int \left(2 x^{3} - 9\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(2 x^{3} - 9\right)d x}}} = {\color{red}{\left(- \int{9 d x} + \int{2 x^{3} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=9$$$:

$$\int{2 x^{3} d x} - {\color{red}{\int{9 d x}}} = \int{2 x^{3} d x} - {\color{red}{\left(9 x\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2$$$ and $$$f{\left(x \right)} = x^{3}$$$:

$$- 9 x + {\color{red}{\int{2 x^{3} d x}}} = - 9 x + {\color{red}{\left(2 \int{x^{3} d x}\right)}}$$

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

$$- 9 x + 2 {\color{red}{\int{x^{3} d x}}}=- 9 x + 2 {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=- 9 x + 2 {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$

Therefore,

$$\int{\left(2 x^{3} - 9\right)d x} = \frac{x^{4}}{2} - 9 x$$

Simplify:

$$\int{\left(2 x^{3} - 9\right)d x} = \frac{x \left(x^{3} - 18\right)}{2}$$

Add the constant of integration:

$$\int{\left(2 x^{3} - 9\right)d x} = \frac{x \left(x^{3} - 18\right)}{2}+C$$

$$$\int \left(2 x^{3} - 9\right)\, dx = \frac{x \left(x^{3} - 18\right)}{2} + C$$$A