Integral of $$$\frac{1}{x^{8}}$$$

Find $$$\int \frac{1}{x^{8}}\, dx$$$.

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-8$$$:

$${\color{red}{\int{\frac{1}{x^{8}} d x}}}={\color{red}{\int{x^{-8} d x}}}={\color{red}{\frac{x^{-8 + 1}}{-8 + 1}}}={\color{red}{\left(- \frac{x^{-7}}{7}\right)}}={\color{red}{\left(- \frac{1}{7 x^{7}}\right)}}$$

Therefore,

$$\int{\frac{1}{x^{8}} d x} = - \frac{1}{7 x^{7}}$$

Add the constant of integration:

$$\int{\frac{1}{x^{8}} d x} = - \frac{1}{7 x^{7}}+C$$

$$$\int \frac{1}{x^{8}}\, dx = - \frac{1}{7 x^{7}} + C$$$A