$$$1 + \frac{1}{x^{9}}$$$ 的積分

求$$$\int \left(1 + \frac{1}{x^{9}}\right)\, dx$$$。

逐項積分：

$${\color{red}{\int{\left(1 + \frac{1}{x^{9}}\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{\frac{1}{x^{9}} d x}\right)}}$$

配合 $$$c=1$$$，應用常數法則 $$$\int c\, dx = c x$$$：

$$\int{\frac{1}{x^{9}} d x} + {\color{red}{\int{1 d x}}} = \int{\frac{1}{x^{9}} d x} + {\color{red}{x}}$$

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-9$$$：

$$x + {\color{red}{\int{\frac{1}{x^{9}} d x}}}=x + {\color{red}{\int{x^{-9} d x}}}=x + {\color{red}{\frac{x^{-9 + 1}}{-9 + 1}}}=x + {\color{red}{\left(- \frac{x^{-8}}{8}\right)}}=x + {\color{red}{\left(- \frac{1}{8 x^{8}}\right)}}$$

因此，

$$\int{\left(1 + \frac{1}{x^{9}}\right)d x} = x - \frac{1}{8 x^{8}}$$

加上積分常數：

$$\int{\left(1 + \frac{1}{x^{9}}\right)d x} = x - \frac{1}{8 x^{8}}+C$$

$$$\int \left(1 + \frac{1}{x^{9}}\right)\, dx = \left(x - \frac{1}{8 x^{8}}\right) + C$$$A