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

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

设$$$u=\ln{\left(x \right)}$$$。

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

积分变为

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

$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

回忆一下 $$$u=\ln{\left(x \right)}$$$:

$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{{\color{red}{\ln{\left(x \right)}}}}\right| \right)}$$

因此，

$$\int{\frac{1}{x \ln{\left(x \right)}} d x} = \ln{\left(\left|{\ln{\left(x \right)}}\right| \right)}$$

加上积分常数：

$$\int{\frac{1}{x \ln{\left(x \right)}} d x} = \ln{\left(\left|{\ln{\left(x \right)}}\right| \right)}+C$$

$$$\int \frac{1}{x \ln\left(x\right)}\, dx = \ln\left(\left|{\ln\left(x\right)}\right|\right) + C$$$A