Integral de $$$\frac{2}{3 x^{4}}$$$

Halla $$$\int \frac{2}{3 x^{4}}\, dx$$$.

Aplica la regla del factor constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ con $$$c=\frac{2}{3}$$$ y $$$f{\left(x \right)} = \frac{1}{x^{4}}$$$:

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

Aplica la regla de la potencia $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ con $$$n=-4$$$:

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

Por lo tanto,

$$\int{\frac{2}{3 x^{4}} d x} = - \frac{2}{9 x^{3}}$$

Añade la constante de integración:

$$\int{\frac{2}{3 x^{4}} d x} = - \frac{2}{9 x^{3}}+C$$

$$$\int \frac{2}{3 x^{4}}\, dx = - \frac{2}{9 x^{3}} + C$$$A