Integral of $$$\frac{1}{t^{\frac{3}{4}}}$$$

Find $$$\int \frac{1}{t^{\frac{3}{4}}}\, dt$$$.

Apply the power rule $$$\int t^{n}\, dt = \frac{t^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=- \frac{3}{4}$$$:

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

Therefore,

$$\int{\frac{1}{t^{\frac{3}{4}}} d t} = 4 \sqrt[4]{t}$$

Add the constant of integration:

$$\int{\frac{1}{t^{\frac{3}{4}}} d t} = 4 \sqrt[4]{t}+C$$

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