$$$\frac{1}{x y^{2}}$$$ 关于$$$x$$$的积分

求$$$\int \frac{1}{x y^{2}}\, dx$$$。

对 $$$c=\frac{1}{y^{2}}$$$ 和 $$$f{\left(x \right)} = \frac{1}{x}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

$$$\frac{1}{x}$$$ 的积分为 $$$\int{\frac{1}{x} d x} = \ln{\left(\left|{x}\right| \right)}$$$:

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

因此，

$$\int{\frac{1}{x y^{2}} d x} = \frac{\ln{\left(\left|{x}\right| \right)}}{y^{2}}$$

加上积分常数：

$$\int{\frac{1}{x y^{2}} d x} = \frac{\ln{\left(\left|{x}\right| \right)}}{y^{2}}+C$$

$$$\int \frac{1}{x y^{2}}\, dx = \frac{\ln\left(\left|{x}\right|\right)}{y^{2}} + C$$$A