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

Find $$$\int \sqrt{2} \sqrt{\frac{1}{x}}\, dx$$$.

The input is rewritten: $$$\int{\sqrt{2} \sqrt{\frac{1}{x}} d x}=\int{\frac{\sqrt{2}}{\sqrt{x}} d x}$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\sqrt{2}$$$ and $$$f{\left(x \right)} = \frac{1}{\sqrt{x}}$$$:

$${\color{red}{\int{\frac{\sqrt{2}}{\sqrt{x}} d x}}} = {\color{red}{\sqrt{2} \int{\frac{1}{\sqrt{x}} d x}}}$$

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

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

Therefore,

$$\int{\frac{\sqrt{2}}{\sqrt{x}} d x} = 2 \sqrt{2} \sqrt{x}$$

Add the constant of integration:

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

$$$\int \sqrt{2} \sqrt{\frac{1}{x}}\, dx = 2 \sqrt{2} \sqrt{x} + C$$$A