Integral of $$$\sqrt{3} z^{32} \left(x - 3\right)$$$ with respect to $$$x$$$

Find $$$\int \sqrt{3} z^{32} \left(x - 3\right)\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\sqrt{3} z^{32}$$$ and $$$f{\left(x \right)} = x - 3$$$:

$${\color{red}{\int{\sqrt{3} z^{32} \left(x - 3\right) d x}}} = {\color{red}{\sqrt{3} z^{32} \int{\left(x - 3\right)d x}}}$$

Integrate term by term:

$$\sqrt{3} z^{32} {\color{red}{\int{\left(x - 3\right)d x}}} = \sqrt{3} z^{32} {\color{red}{\left(- \int{3 d x} + \int{x d x}\right)}}$$

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

$$\sqrt{3} z^{32} \left(\int{x d x} - {\color{red}{\int{3 d x}}}\right) = \sqrt{3} z^{32} \left(\int{x d x} - {\color{red}{\left(3 x\right)}}\right)$$

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

$$\sqrt{3} z^{32} \left(- 3 x + {\color{red}{\int{x d x}}}\right)=\sqrt{3} z^{32} \left(- 3 x + {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}\right)=\sqrt{3} z^{32} \left(- 3 x + {\color{red}{\left(\frac{x^{2}}{2}\right)}}\right)$$

Therefore,

$$\int{\sqrt{3} z^{32} \left(x - 3\right) d x} = \sqrt{3} z^{32} \left(\frac{x^{2}}{2} - 3 x\right)$$

Simplify:

$$\int{\sqrt{3} z^{32} \left(x - 3\right) d x} = \frac{\sqrt{3} x z^{32} \left(x - 6\right)}{2}$$

Add the constant of integration:

$$\int{\sqrt{3} z^{32} \left(x - 3\right) d x} = \frac{\sqrt{3} x z^{32} \left(x - 6\right)}{2}+C$$

$$$\int \sqrt{3} z^{32} \left(x - 3\right)\, dx = \frac{\sqrt{3} x z^{32} \left(x - 6\right)}{2} + C$$$A