Integral de $$$- \frac{1}{4 x^{3}}$$$

Halla $$$\int \left(- \frac{1}{4 x^{3}}\right)\, dx$$$.

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=- \frac{1}{4}$$$ y $$$f{\left(x \right)} = \frac{1}{x^{3}}$$$:

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

Aplica la regla de la potencia $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=-3$$$:

$$- \frac{{\color{red}{\int{\frac{1}{x^{3}} d x}}}}{4}=- \frac{{\color{red}{\int{x^{-3} d x}}}}{4}=- \frac{{\color{red}{\frac{x^{-3 + 1}}{-3 + 1}}}}{4}=- \frac{{\color{red}{\left(- \frac{x^{-2}}{2}\right)}}}{4}=- \frac{{\color{red}{\left(- \frac{1}{2 x^{2}}\right)}}}{4}$$

Por lo tanto,

$$\int{\left(- \frac{1}{4 x^{3}}\right)d x} = \frac{1}{8 x^{2}}$$

Añade la constante de integración:

$$\int{\left(- \frac{1}{4 x^{3}}\right)d x} = \frac{1}{8 x^{2}}+C$$

$$$\int \left(- \frac{1}{4 x^{3}}\right)\, dx = \frac{1}{8 x^{2}} + C$$$A