Integral of $$$\frac{\sqrt{2} x}{2 \sqrt{x^{3}}}$$$

Find $$$\int \frac{\sqrt{2} x}{2 \sqrt{x^{3}}}\, dx$$$.

The input is rewritten: $$$\int{\frac{\sqrt{2} x}{2 \sqrt{x^{3}}} d x}=\int{\frac{\sqrt{2}}{2 \sqrt{x}} d x}$$$.

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

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

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

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

Therefore,

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

Add the constant of integration:

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

$$$\int \frac{\sqrt{2} x}{2 \sqrt{x^{3}}}\, dx = \sqrt{2} \sqrt{x} + C$$$A