$$$\ln\left(x_{0}\right)$$$ 的积分

求$$$\int \ln\left(x_{0}\right)\, dx_{0}$$$。

对于积分$$$\int{\ln{\left(x_{0} \right)} d x_{0}}$$$，使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

设 $$$\operatorname{u}=\ln{\left(x_{0} \right)}$$$ 和 $$$\operatorname{dv}=dx_{0}$$$。

则 $$$\operatorname{du}=\left(\ln{\left(x_{0} \right)}\right)^{\prime }dx_{0}=\frac{dx_{0}}{x_{0}}$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{1 d x_{0}}=x_{0}$$$ (步骤见 »)。

因此，

$${\color{red}{\int{\ln{\left(x_{0} \right)} d x_{0}}}}={\color{red}{\left(\ln{\left(x_{0} \right)} \cdot x_{0}-\int{x_{0} \cdot \frac{1}{x_{0}} d x_{0}}\right)}}={\color{red}{\left(x_{0} \ln{\left(x_{0} \right)} - \int{1 d x_{0}}\right)}}$$

应用常数法则 $$$\int c\, dx_{0} = c x_{0}$$$，使用 $$$c=1$$$：

$$x_{0} \ln{\left(x_{0} \right)} - {\color{red}{\int{1 d x_{0}}}} = x_{0} \ln{\left(x_{0} \right)} - {\color{red}{x_{0}}}$$

因此，

$$\int{\ln{\left(x_{0} \right)} d x_{0}} = x_{0} \ln{\left(x_{0} \right)} - x_{0}$$

化简：

$$\int{\ln{\left(x_{0} \right)} d x_{0}} = x_{0} \left(\ln{\left(x_{0} \right)} - 1\right)$$

加上积分常数：

$$\int{\ln{\left(x_{0} \right)} d x_{0}} = x_{0} \left(\ln{\left(x_{0} \right)} - 1\right)+C$$

$$$\int \ln\left(x_{0}\right)\, dx_{0} = x_{0} \left(\ln\left(x_{0}\right) - 1\right) + C$$$A