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

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

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

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

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

Therefore,

$$\int{\sqrt{2} \sqrt{x} d x} = \frac{2 \sqrt{2} x^{\frac{3}{2}}}{3}$$

Add the constant of integration:

$$\int{\sqrt{2} \sqrt{x} d x} = \frac{2 \sqrt{2} x^{\frac{3}{2}}}{3}+C$$

Answer: $$$\int{\sqrt{2} \sqrt{x} d x}=\frac{2 \sqrt{2} x^{\frac{3}{2}}}{3}+C$$$