$$$\ln\left(x_{0}\right)$$$ 的積分

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

對於積分 $$$\int{\ln{\left(x_{0} \right)} d x_{0}}$$$，使用分部積分法 $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

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

則 $$$\operatorname{du}=\left(\ln{\left(x_{0} \right)}\right)^{\prime }dx_{0}=\frac{dx_{0}}{x_{0}}$$$（步驟見 »），且 $$$\operatorname{v}=\int{1 d x_{0}}=x_{0}$$$（步驟見 »）。

因此，

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

配合 $$$c=1$$$，應用常數法則 $$$\int c\, dx_{0} = c x_{0}$$$：

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

因此，

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

化簡：

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

加上積分常數：

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

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