Integral of $$$- \frac{1}{2 t^{\frac{4}{3}}}$$$

Find $$$\int \left(- \frac{1}{2 t^{\frac{4}{3}}}\right)\, dt$$$.

Apply the constant multiple rule $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ with $$$c=- \frac{1}{2}$$$ and $$$f{\left(t \right)} = \frac{1}{t^{\frac{4}{3}}}$$$:

$${\color{red}{\int{\left(- \frac{1}{2 t^{\frac{4}{3}}}\right)d t}}} = {\color{red}{\left(- \frac{\int{\frac{1}{t^{\frac{4}{3}}} d t}}{2}\right)}}$$

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

$$- \frac{{\color{red}{\int{\frac{1}{t^{\frac{4}{3}}} d t}}}}{2}=- \frac{{\color{red}{\int{t^{- \frac{4}{3}} d t}}}}{2}=- \frac{{\color{red}{\frac{t^{- \frac{4}{3} + 1}}{- \frac{4}{3} + 1}}}}{2}=- \frac{{\color{red}{\left(- 3 t^{- \frac{1}{3}}\right)}}}{2}=- \frac{{\color{red}{\left(- \frac{3}{\sqrt[3]{t}}\right)}}}{2}$$

Therefore,

$$\int{\left(- \frac{1}{2 t^{\frac{4}{3}}}\right)d t} = \frac{3}{2 \sqrt[3]{t}}$$

Add the constant of integration:

$$\int{\left(- \frac{1}{2 t^{\frac{4}{3}}}\right)d t} = \frac{3}{2 \sqrt[3]{t}}+C$$

$$$\int \left(- \frac{1}{2 t^{\frac{4}{3}}}\right)\, dt = \frac{3}{2 \sqrt[3]{t}} + C$$$A