Integral de $$$- \frac{2}{t^{3}}$$$

Encontre $$$\int \left(- \frac{2}{t^{3}}\right)\, dt$$$.

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

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

Aplique a regra da potência $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=-3$$$:

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

Portanto,

$$\int{\left(- \frac{2}{t^{3}}\right)d t} = \frac{1}{t^{2}}$$

Adicione a constante de integração:

$$\int{\left(- \frac{2}{t^{3}}\right)d t} = \frac{1}{t^{2}}+C$$

$$$\int \left(- \frac{2}{t^{3}}\right)\, dt = \frac{1}{t^{2}} + C$$$A