Integral of $$$\frac{x^{3}}{z} - 2$$$ with respect to $$$x$$$

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

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=2$$$:

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{z}$$$ and $$$f{\left(x \right)} = x^{3}$$$:

$$- 2 x + {\color{red}{\int{\frac{x^{3}}{z} d x}}} = - 2 x + {\color{red}{\frac{\int{x^{3} d x}}{z}}}$$

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

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

Therefore,

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

Add the constant of integration:

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

$$$\int \left(\frac{x^{3}}{z} - 2\right)\, dx = \left(\frac{x^{4}}{4 z} - 2 x\right) + C$$$A