Integral of $$$-8 + \frac{1}{x^{3}}$$$

Find $$$\int \left(-8 + \frac{1}{x^{3}}\right)\, dx$$$.

Integrate term by term:

$${\color{red}{\int{\left(-8 + \frac{1}{x^{3}}\right)d x}}} = {\color{red}{\left(- \int{8 d x} + \int{\frac{1}{x^{3}} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=8$$$:

$$\int{\frac{1}{x^{3}} d x} - {\color{red}{\int{8 d x}}} = \int{\frac{1}{x^{3}} d x} - {\color{red}{\left(8 x\right)}}$$

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

$$- 8 x + {\color{red}{\int{\frac{1}{x^{3}} d x}}}=- 8 x + {\color{red}{\int{x^{-3} d x}}}=- 8 x + {\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}}=- 8 x + {\color{red}{\left(- \frac{x^{-2}}{2}\right)}}=- 8 x + {\color{red}{\left(- \frac{1}{2 x^{2}}\right)}}$$

Therefore,

$$\int{\left(-8 + \frac{1}{x^{3}}\right)d x} = - 8 x - \frac{1}{2 x^{2}}$$

Add the constant of integration:

$$\int{\left(-8 + \frac{1}{x^{3}}\right)d x} = - 8 x - \frac{1}{2 x^{2}}+C$$

$$$\int \left(-8 + \frac{1}{x^{3}}\right)\, dx = \left(- 8 x - \frac{1}{2 x^{2}}\right) + C$$$A