Integral de $$$\frac{1}{3 u^{4}}$$$

Halla $$$\int \frac{1}{3 u^{4}}\, du$$$.

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

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

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

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

Por lo tanto,

$$\int{\frac{1}{3 u^{4}} d u} = - \frac{1}{9 u^{3}}$$

Añade la constante de integración:

$$\int{\frac{1}{3 u^{4}} d u} = - \frac{1}{9 u^{3}}+C$$

$$$\int \frac{1}{3 u^{4}}\, du = - \frac{1}{9 u^{3}} + C$$$A