$$$\frac{1}{x \ln^{9}\left(x\right)}$$$ 的积分

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

设$$$u=\ln{\left(x \right)}$$$。

则$$$du=\left(\ln{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x}$$$ (步骤见»)，并有$$$\frac{dx}{x} = du$$$。

该积分可以改写为

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

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=-9$$$：

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

回忆一下 $$$u=\ln{\left(x \right)}$$$:

$$- \frac{{\color{red}{u}}^{-8}}{8} = - \frac{{\color{red}{\ln{\left(x \right)}}}^{-8}}{8}$$

因此，

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

加上积分常数：

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

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