Integral of $$$\frac{\sqrt{x y}}{x^{2} y^{2}}$$$ with respect to $$$x$$$

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

The input is rewritten: $$$\int{\frac{\sqrt{x y}}{x^{2} y^{2}} d x}=\int{\frac{1}{x^{\frac{3}{2}} y^{\frac{3}{2}}} d x}$$$.

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{y^{\frac{3}{2}}}$$$ and $$$f{\left(x \right)} = \frac{1}{x^{\frac{3}{2}}}$$$:

$${\color{red}{\int{\frac{1}{x^{\frac{3}{2}} y^{\frac{3}{2}}} d x}}} = {\color{red}{\frac{\int{\frac{1}{x^{\frac{3}{2}}} d x}}{y^{\frac{3}{2}}}}}$$

Apply the power rule $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ with $$$n=- \frac{3}{2}$$$:

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

Therefore,

$$\int{\frac{1}{x^{\frac{3}{2}} y^{\frac{3}{2}}} d x} = - \frac{2}{\sqrt{x} y^{\frac{3}{2}}}$$

Add the constant of integration:

$$\int{\frac{1}{x^{\frac{3}{2}} y^{\frac{3}{2}}} d x} = - \frac{2}{\sqrt{x} y^{\frac{3}{2}}}+C$$

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