Integral of $$$\frac{1}{2 u^{3}}$$$

Find $$$\int \frac{1}{2 u^{3}}\, du$$$.

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

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

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

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

Therefore,

$$\int{\frac{1}{2 u^{3}} d u} = - \frac{1}{4 u^{2}}$$

Add the constant of integration:

$$\int{\frac{1}{2 u^{3}} d u} = - \frac{1}{4 u^{2}}+C$$

$$$\int \frac{1}{2 u^{3}}\, du = - \frac{1}{4 u^{2}} + C$$$A