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

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

Aplique a regra do múltiplo constante $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ usando $$$c=\frac{2}{3}$$$ e $$$f{\left(x \right)} = \frac{1}{x^{4}}$$$:

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

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

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

Portanto,

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

Adicione a constante de integração:

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

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