Integral of $$$\frac{1}{t^{6}}$$$

Find $$$\int \frac{1}{t^{6}}\, dt$$$.

Apply the power rule $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-6$$$:

$${\color{red}{\int{\frac{1}{t^{6}} d t}}}={\color{red}{\int{t^{-6} d t}}}={\color{red}{\frac{t^{-6 + 1}}{-6 + 1}}}={\color{red}{\left(- \frac{t^{-5}}{5}\right)}}={\color{red}{\left(- \frac{1}{5 t^{5}}\right)}}$$

Therefore,

$$\int{\frac{1}{t^{6}} d t} = - \frac{1}{5 t^{5}}$$

Add the constant of integration:

$$\int{\frac{1}{t^{6}} d t} = - \frac{1}{5 t^{5}}+C$$

$$$\int \frac{1}{t^{6}}\, dt = - \frac{1}{5 t^{5}} + C$$$A