Integral of $$$\frac{6 x^{2} - 1}{x^{2}}$$$

Find $$$\int \frac{6 x^{2} - 1}{x^{2}}\, dx$$$.

Expand the expression:

$${\color{red}{\int{\frac{6 x^{2} - 1}{x^{2}} d x}}} = {\color{red}{\int{\left(6 - \frac{1}{x^{2}}\right)d x}}}$$

Integrate term by term:

$${\color{red}{\int{\left(6 - \frac{1}{x^{2}}\right)d x}}} = {\color{red}{\left(\int{6 d x} - \int{\frac{1}{x^{2}} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=6$$$:

$$- \int{\frac{1}{x^{2}} d x} + {\color{red}{\int{6 d x}}} = - \int{\frac{1}{x^{2}} d x} + {\color{red}{\left(6 x\right)}}$$

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

$$6 x - {\color{red}{\int{\frac{1}{x^{2}} d x}}}=6 x - {\color{red}{\int{x^{-2} d x}}}=6 x - {\color{red}{\frac{x^{-2 + 1}}{-2 + 1}}}=6 x - {\color{red}{\left(- x^{-1}\right)}}=6 x - {\color{red}{\left(- \frac{1}{x}\right)}}$$

Therefore,

$$\int{\frac{6 x^{2} - 1}{x^{2}} d x} = 6 x + \frac{1}{x}$$

Add the constant of integration:

$$\int{\frac{6 x^{2} - 1}{x^{2}} d x} = 6 x + \frac{1}{x}+C$$

$$$\int \frac{6 x^{2} - 1}{x^{2}}\, dx = \left(6 x + \frac{1}{x}\right) + C$$$A