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

Find $$$\int \left(-1 + \frac{1}{x^{2}}\right)\, dx$$$.

Integrate term by term:

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

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:

$$\int{\frac{1}{x^{2}} d x} - {\color{red}{\int{1 d x}}} = \int{\frac{1}{x^{2}} d x} - {\color{red}{x}}$$

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

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

Therefore,

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

Add the constant of integration:

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

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