Integral of $$$\frac{1}{x^{4}}$$$

Find $$$\int \frac{1}{x^{4}}\, dx$$$.

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-4$$$:

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

Therefore,

$$\int{\frac{1}{x^{4}} d x} = - \frac{1}{3 x^{3}}$$

Add the constant of integration:

$$\int{\frac{1}{x^{4}} d x} = - \frac{1}{3 x^{3}}+C$$

$$$\int \frac{1}{x^{4}}\, dx = - \frac{1}{3 x^{3}} + C$$$A