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

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

对于积分$$$\int{\frac{\ln{\left(x \right)}}{x^{2}} d x}$$$，使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

设 $$$\operatorname{u}=\ln{\left(x \right)}$$$ 和 $$$\operatorname{dv}=\frac{dx}{x^{2}}$$$。

则 $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{\frac{1}{x^{2}} d x}=- \frac{1}{x}$$$ (步骤见 »)。

该积分可以改写为

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

对 $$$c=-1$$$ 和 $$$f{\left(x \right)} = \frac{1}{x^{2}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

应用幂法则 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=-2$$$：

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

因此，

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

化简：

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

加上积分常数：

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

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