$$$\frac{x}{\ln\left(x\right)}$$$ 的积分

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

设$$$u=\ln{\left(x \right)}$$$。

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

该积分可以改写为

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

设$$$v=2 u$$$。

则$$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (步骤见»)，并有$$$du = \frac{dv}{2}$$$。

积分变为

$${\color{red}{\int{\frac{e^{2 u}}{u} d u}}} = {\color{red}{\int{\frac{e^{v}}{v} d v}}}$$

该积分（指数积分）没有闭式表达式：

$${\color{red}{\int{\frac{e^{v}}{v} d v}}} = {\color{red}{\operatorname{Ei}{\left(v \right)}}}$$

回忆一下 $$$v=2 u$$$:

$$\operatorname{Ei}{\left({\color{red}{v}} \right)} = \operatorname{Ei}{\left({\color{red}{\left(2 u\right)}} \right)}$$

回忆一下 $$$u=\ln{\left(x \right)}$$$:

$$\operatorname{Ei}{\left(2 {\color{red}{u}} \right)} = \operatorname{Ei}{\left(2 {\color{red}{\ln{\left(x \right)}}} \right)}$$

因此，

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

加上积分常数：

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

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