Integral of $$$\frac{1}{x^{3}}$$$

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

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

Therefore,

$$\int{\frac{1}{x^{3}} d x} = - \frac{1}{2 x^{2}}$$

Add the constant of integration:

$$\int{\frac{1}{x^{3}} d x} = - \frac{1}{2 x^{2}}+C$$

$$$\int \frac{1}{x^{3}}\, dx = - \frac{1}{2 x^{2}} + C$$$A