Integral of $$$4 x^{3} - 3 x^{2} - 8 x$$$

Find $$$\int \left(4 x^{3} - 3 x^{2} - 8 x\right)\, dx$$$.

Integrate term by term:

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=8$$$ and $$$f{\left(x \right)} = x$$$:

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

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

$$- \int{3 x^{2} d x} + \int{4 x^{3} d x} - 8 {\color{red}{\int{x d x}}}=- \int{3 x^{2} d x} + \int{4 x^{3} d x} - 8 {\color{red}{\frac{x^{1 + 1}}{1 + 1}}}=- \int{3 x^{2} d x} + \int{4 x^{3} d x} - 8 {\color{red}{\left(\frac{x^{2}}{2}\right)}}$$

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

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

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=2$$$:

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = x^{3}$$$:

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

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

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

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