Integral of $$$\frac{4 x^{4} - 15 x^{3}}{x}$$$

Find $$$\int \frac{4 x^{4} - 15 x^{3}}{x}\, dx$$$.

Expand the expression:

$${\color{red}{\int{\frac{4 x^{4} - 15 x^{3}}{x} d x}}} = {\color{red}{\int{\left(4 x^{3} - 15 x^{2}\right)d x}}}$$

Integrate term by term:

$${\color{red}{\int{\left(4 x^{3} - 15 x^{2}\right)d x}}} = {\color{red}{\left(- \int{15 x^{2} d x} + \int{4 x^{3} d x}\right)}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=15$$$ and $$$f{\left(x \right)} = x^{2}$$$:

$$\int{4 x^{3} d x} - {\color{red}{\int{15 x^{2} d x}}} = \int{4 x^{3} d x} - {\color{red}{\left(15 \int{x^{2} d x}\right)}}$$

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

$$\int{4 x^{3} d x} - 15 {\color{red}{\int{x^{2} d x}}}=\int{4 x^{3} d x} - 15 {\color{red}{\frac{x^{1 + 2}}{1 + 2}}}=\int{4 x^{3} d x} - 15 {\color{red}{\left(\frac{x^{3}}{3}\right)}}$$

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

$$- 5 x^{3} + {\color{red}{\int{4 x^{3} d x}}} = - 5 x^{3} + {\color{red}{\left(4 \int{x^{3} d x}\right)}}$$

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

$$- 5 x^{3} + 4 {\color{red}{\int{x^{3} d x}}}=- 5 x^{3} + 4 {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=- 5 x^{3} + 4 {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$

Therefore,

$$\int{\frac{4 x^{4} - 15 x^{3}}{x} d x} = x^{4} - 5 x^{3}$$

Simplify:

$$\int{\frac{4 x^{4} - 15 x^{3}}{x} d x} = x^{3} \left(x - 5\right)$$

Add the constant of integration:

$$\int{\frac{4 x^{4} - 15 x^{3}}{x} d x} = x^{3} \left(x - 5\right)+C$$

$$$\int \frac{4 x^{4} - 15 x^{3}}{x}\, dx = x^{3} \left(x - 5\right) + C$$$A