Integral de $$$\frac{1}{z^{4}}$$$

Encontre $$$\int \frac{1}{z^{4}}\, dz$$$.

Aplique a regra da potência $$$\int z^{n}\, dz = \frac{z^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=-4$$$:

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

Portanto,

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

Adicione a constante de integração:

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

$$$\int \frac{1}{z^{4}}\, dz = - \frac{1}{3 z^{3}} + C$$$A