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

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

Integrate term by term:

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

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

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

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

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

Therefore,

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

Simplify:

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

Add the constant of integration:

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

$$$\int \left(- x^{2} + \frac{1}{x^{2}}\right)\, dx = \frac{- x^{4} - 3}{3 x} + C$$$A