Integral of $$$\frac{1}{t^{4}}$$$

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

Apply the power rule $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=-4$$$:

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

Therefore,

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

Add the constant of integration:

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

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