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

Find $$$\int \frac{1}{e x^{3}}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=e^{-1}$$$ and $$$f{\left(x \right)} = \frac{1}{x^{3}}$$$:

$${\color{red}{\int{\frac{1}{e x^{3}} d x}}} = {\color{red}{\frac{\int{\frac{1}{x^{3}} d x}}{e}}}$$

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

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

Therefore,

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

Add the constant of integration:

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

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