$$$\eta n - x^{3}$$$ 关于$$$x$$$的积分

求$$$\int \left(\eta n - x^{3}\right)\, dx$$$。

逐项积分：

$${\color{red}{\int{\left(\eta n - x^{3}\right)d x}}} = {\color{red}{\left(- \int{x^{3} d x} + \int{\eta n d x}\right)}}$$

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=3$$$：

$$\int{\eta n d x} - {\color{red}{\int{x^{3} d x}}}=\int{\eta n d x} - {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=\int{\eta n d x} - {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=\eta n$$$：

$$- \frac{x^{4}}{4} + {\color{red}{\int{\eta n d x}}} = - \frac{x^{4}}{4} + {\color{red}{\eta n x}}$$

因此，

$$\int{\left(\eta n - x^{3}\right)d x} = \eta n x - \frac{x^{4}}{4}$$

化简：

$$\int{\left(\eta n - x^{3}\right)d x} = x \left(\eta n - \frac{x^{3}}{4}\right)$$

加上积分常数：

$$\int{\left(\eta n - x^{3}\right)d x} = x \left(\eta n - \frac{x^{3}}{4}\right)+C$$

$$$\int \left(\eta n - x^{3}\right)\, dx = x \left(\eta n - \frac{x^{3}}{4}\right) + C$$$A