Integral of $$$- x^{2} + \frac{1}{x^{3}}$$$

Find $$$\int \left(- x^{2} + \frac{1}{x^{3}}\right)\, dx$$$.

Integrate term by term:

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

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

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

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

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

Therefore,

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

Simplify:

$$\int{\left(- x^{2} + \frac{1}{x^{3}}\right)d x} = \frac{- 2 x^{5} - 3}{6 x^{2}}$$

Add the constant of integration:

$$\int{\left(- x^{2} + \frac{1}{x^{3}}\right)d x} = \frac{- 2 x^{5} - 3}{6 x^{2}}+C$$

$$$\int \left(- x^{2} + \frac{1}{x^{3}}\right)\, dx = \frac{- 2 x^{5} - 3}{6 x^{2}} + C$$$A