$$$\frac{1}{x^{2} y}$$$ 关于$$$y$$$的积分

求$$$\int \frac{1}{x^{2} y}\, dy$$$。

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

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

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

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

因此，

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

加上积分常数：

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

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