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

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

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

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

Por lo tanto,

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

Añade la constante de integración:

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

$$$\int \frac{1}{u^{3}}\, du = - \frac{1}{2 u^{2}} + C$$$A