$$$\eta n - x^{3}$$$ 對 $$$x$$$ 的積分
您的輸入
求$$$\int \left(\eta n - x^{3}\right)\, dx$$$。
解答
逐項積分：
$${\color{red}{\int{\left(\eta n - x^{3}\right)d x}}} = {\color{red}{\left(- \int{x^{3} d x} + \int{\eta n d x}\right)}}$$
套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=3$$$：
$$\int{\eta n d x} - {\color{red}{\int{x^{3} d x}}}=\int{\eta n d x} - {\color{red}{\frac{x^{1 + 3}}{1 + 3}}}=\int{\eta n d x} - {\color{red}{\left(\frac{x^{4}}{4}\right)}}$$
配合 $$$c=\eta n$$$，應用常數法則 $$$\int c\, dx = c x$$$：
$$- \frac{x^{4}}{4} + {\color{red}{\int{\eta n d x}}} = - \frac{x^{4}}{4} + {\color{red}{\eta n x}}$$
因此，
$$\int{\left(\eta n - x^{3}\right)d x} = \eta n x - \frac{x^{4}}{4}$$
化簡：
$$\int{\left(\eta n - x^{3}\right)d x} = x \left(\eta n - \frac{x^{3}}{4}\right)$$
加上積分常數：
$$\int{\left(\eta n - x^{3}\right)d x} = x \left(\eta n - \frac{x^{3}}{4}\right)+C$$
答案
$$$\int \left(\eta n - x^{3}\right)\, dx = x \left(\eta n - \frac{x^{3}}{4}\right) + C$$$A