Integral of $$$\frac{x^{4} - 4}{x^{2}}$$$

Find $$$\int \frac{x^{4} - 4}{x^{2}}\, dx$$$.

Expand the expression:

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

Integrate term by term:

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

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

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

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=4$$$ and $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$:

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

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

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

Therefore,

$$\int{\frac{x^{4} - 4}{x^{2}} d x} = \frac{x^{3}}{3} + \frac{4}{x}$$

Simplify:

$$\int{\frac{x^{4} - 4}{x^{2}} d x} = \frac{x^{4} + 12}{3 x}$$

Add the constant of integration:

$$\int{\frac{x^{4} - 4}{x^{2}} d x} = \frac{x^{4} + 12}{3 x}+C$$

$$$\int \frac{x^{4} - 4}{x^{2}}\, dx = \frac{x^{4} + 12}{3 x} + C$$$A