$$$- 10 a f n^{2} t^{2} x^{21} y$$$ 关于$$$x$$$的积分

求$$$\int \left(- 10 a f n^{2} t^{2} x^{21} y\right)\, dx$$$。

对 $$$c=- 10 a f n^{2} t^{2} y$$$ 和 $$$f{\left(x \right)} = x^{21}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\left(- 10 a f n^{2} t^{2} x^{21} y\right)d x}}} = {\color{red}{\left(- 10 a f n^{2} t^{2} y \int{x^{21} d x}\right)}}$$

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=21$$$：

$$- 10 a f n^{2} t^{2} y {\color{red}{\int{x^{21} d x}}}=- 10 a f n^{2} t^{2} y {\color{red}{\frac{x^{1 + 21}}{1 + 21}}}=- 10 a f n^{2} t^{2} y {\color{red}{\left(\frac{x^{22}}{22}\right)}}$$

因此，

$$\int{\left(- 10 a f n^{2} t^{2} x^{21} y\right)d x} = - \frac{5 a f n^{2} t^{2} x^{22} y}{11}$$

加上积分常数：

$$\int{\left(- 10 a f n^{2} t^{2} x^{21} y\right)d x} = - \frac{5 a f n^{2} t^{2} x^{22} y}{11}+C$$

$$$\int \left(- 10 a f n^{2} t^{2} x^{21} y\right)\, dx = - \frac{5 a f n^{2} t^{2} x^{22} y}{11} + C$$$A