$$$1 + \frac{1}{x^{5}}$$$ 的積分

求$$$\int \left(1 + \frac{1}{x^{5}}\right)\, dx$$$。

逐項積分：

$${\color{red}{\int{\left(1 + \frac{1}{x^{5}}\right)d x}}} = {\color{red}{\left(\int{1 d x} + \int{\frac{1}{x^{5}} d x}\right)}}$$

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

$$\int{\frac{1}{x^{5}} d x} + {\color{red}{\int{1 d x}}} = \int{\frac{1}{x^{5}} d x} + {\color{red}{x}}$$

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-5$$$：

$$x + {\color{red}{\int{\frac{1}{x^{5}} d x}}}=x + {\color{red}{\int{x^{-5} d x}}}=x + {\color{red}{\frac{x^{-5 + 1}}{-5 + 1}}}=x + {\color{red}{\left(- \frac{x^{-4}}{4}\right)}}=x + {\color{red}{\left(- \frac{1}{4 x^{4}}\right)}}$$

因此，

$$\int{\left(1 + \frac{1}{x^{5}}\right)d x} = x - \frac{1}{4 x^{4}}$$

加上積分常數：

$$\int{\left(1 + \frac{1}{x^{5}}\right)d x} = x - \frac{1}{4 x^{4}}+C$$

$$$\int \left(1 + \frac{1}{x^{5}}\right)\, dx = \left(x - \frac{1}{4 x^{4}}\right) + C$$$A