$$$\frac{s^{2}}{d t}$$$ 关于$$$t$$$的积分

求$$$\int \frac{s^{2}}{d t}\, dt$$$。

对 $$$c=\frac{s^{2}}{d}$$$ 和 $$$f{\left(t \right)} = \frac{1}{t}$$$ 应用常数倍法则 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$：

$${\color{red}{\int{\frac{s^{2}}{d t} d t}}} = {\color{red}{\frac{s^{2} \int{\frac{1}{t} d t}}{d}}}$$

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

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

因此，

$$\int{\frac{s^{2}}{d t} d t} = \frac{s^{2} \ln{\left(\left|{t}\right| \right)}}{d}$$

加上积分常数：

$$\int{\frac{s^{2}}{d t} d t} = \frac{s^{2} \ln{\left(\left|{t}\right| \right)}}{d}+C$$

$$$\int \frac{s^{2}}{d t}\, dt = \frac{s^{2} \ln\left(\left|{t}\right|\right)}{d} + C$$$A