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

Find $$$\int \left(- \frac{1}{x^{4}}\right)\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=-1$$$ and $$$f{\left(x \right)} = \frac{1}{x^{4}}$$$:

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

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,

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

Add the constant of integration:

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

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