Integral of $$$\frac{1}{x^{2} y^{2}}$$$ with respect to $$$x$$$

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

$$$\int \frac{1}{x^{2} y^{2}}\, dx = - \frac{1}{x y^{2}} + C$$$A