$$$\frac{1}{x^{2} \ln\left(x\right)}$$$ 的积分

求$$$\int \frac{1}{x^{2} \ln\left(x\right)}\, dx$$$。

设$$$u=\frac{1}{x}$$$。

则$$$du=\left(\frac{1}{x}\right)^{\prime }dx = - \frac{1}{x^{2}} dx$$$ (步骤见»)，并有$$$\frac{dx}{x^{2}} = - du$$$。

所以，

$${\color{red}{\int{\frac{1}{x^{2} \ln{\left(x \right)}} d x}}} = {\color{red}{\int{\frac{1}{\ln{\left(u \right)}} d u}}}$$

该积分（对数积分）没有闭式表达式：

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

回忆一下 $$$u=\frac{1}{x}$$$:

$$\operatorname{li}{\left({\color{red}{u}} \right)} = \operatorname{li}{\left({\color{red}{\frac{1}{x}}} \right)}$$

因此，

$$\int{\frac{1}{x^{2} \ln{\left(x \right)}} d x} = \operatorname{li}{\left(\frac{1}{x} \right)}$$

加上积分常数：

$$\int{\frac{1}{x^{2} \ln{\left(x \right)}} d x} = \operatorname{li}{\left(\frac{1}{x} \right)}+C$$

$$$\int \frac{1}{x^{2} \ln\left(x\right)}\, dx = \operatorname{li}{\left(\frac{1}{x} \right)} + C$$$A