Integral of $$$\frac{3}{\sqrt{x}}$$$

Find $$$\int \frac{3}{\sqrt{x}}\, dx$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=3$$$ and $$$f{\left(x \right)} = \frac{1}{\sqrt{x}}$$$:

$${\color{red}{\int{\frac{3}{\sqrt{x}} d x}}} = {\color{red}{\left(3 \int{\frac{1}{\sqrt{x}} d x}\right)}}$$

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

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

Therefore,

$$\int{\frac{3}{\sqrt{x}} d x} = 6 \sqrt{x}$$

Add the constant of integration:

$$\int{\frac{3}{\sqrt{x}} d x} = 6 \sqrt{x}+C$$

$$$\int \frac{3}{\sqrt{x}}\, dx = 6 \sqrt{x} + C$$$A