Integral of $$$x^{3} - 4 x$$$

Find $$$\int \left(x^{3} - 4 x\right)\, dx$$$.

Integrate term by term:

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

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

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = x$$$:

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

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

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

Therefore,

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

Simplify:

$$\int{\left(x^{3} - 4 x\right)d x} = \frac{x^{2} \left(x^{2} - 8\right)}{4}$$

Add the constant of integration:

$$\int{\left(x^{3} - 4 x\right)d x} = \frac{x^{2} \left(x^{2} - 8\right)}{4}+C$$

$$$\int \left(x^{3} - 4 x\right)\, dx = \frac{x^{2} \left(x^{2} - 8\right)}{4} + C$$$A