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

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

逐项积分：

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

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 »)。

因此，

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

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

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

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

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

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

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

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

因此，

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

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

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

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

设 $$$\operatorname{u}=x$$$ 和 $$$\operatorname{dv}=\sin{\left(x \right)} dx$$$。

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

所以，

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

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

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

余弦函数的积分为 $$$\int{\cos{\left(x \right)} d x} = \sin{\left(x \right)}$$$：

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

因此，

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

化简：

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

加上积分常数：

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

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