Integral de $$$\frac{1}{4 t^{8}}$$$

Encontre $$$\int \frac{1}{4 t^{8}}\, dt$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ usando $$$c=\frac{1}{4}$$$ e $$$f{\left(t \right)} = \frac{1}{t^{8}}$$$:

$${\color{red}{\int{\frac{1}{4 t^{8}} d t}}} = {\color{red}{\left(\frac{\int{\frac{1}{t^{8}} d t}}{4}\right)}}$$

Aplique a regra da potência $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=-8$$$:

$$\frac{{\color{red}{\int{\frac{1}{t^{8}} d t}}}}{4}=\frac{{\color{red}{\int{t^{-8} d t}}}}{4}=\frac{{\color{red}{\frac{t^{-8 + 1}}{-8 + 1}}}}{4}=\frac{{\color{red}{\left(- \frac{t^{-7}}{7}\right)}}}{4}=\frac{{\color{red}{\left(- \frac{1}{7 t^{7}}\right)}}}{4}$$

Portanto,

$$\int{\frac{1}{4 t^{8}} d t} = - \frac{1}{28 t^{7}}$$

Adicione a constante de integração:

$$\int{\frac{1}{4 t^{8}} d t} = - \frac{1}{28 t^{7}}+C$$

$$$\int \frac{1}{4 t^{8}}\, dt = - \frac{1}{28 t^{7}} + C$$$A