Integral of $$$\frac{1}{2 \sqrt{u}}$$$

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

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 $$$f{\left(u \right)} = \frac{1}{\sqrt{u}}$$$:

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

Apply the power rule $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=- \frac{1}{2}$$$:

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

Therefore,

$$\int{\frac{1}{2 \sqrt{u}} d u} = \sqrt{u}$$

Add the constant of integration:

$$\int{\frac{1}{2 \sqrt{u}} d u} = \sqrt{u}+C$$

$$$\int \frac{1}{2 \sqrt{u}}\, du = \sqrt{u} + C$$$A