$$$\frac{\ln^{4}\left(x\right)}{2}$$$ 的积分

求$$$\int \frac{\ln^{4}\left(x\right)}{2}\, dx$$$。

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = \ln{\left(x \right)}^{4}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\frac{\ln{\left(x \right)}^{4}}{2} d x}}} = {\color{red}{\left(\frac{\int{\ln{\left(x \right)}^{4} d x}}{2}\right)}}$$

对于积分$$$\int{\ln{\left(x \right)}^{4} d x}$$$，使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

设 $$$\operatorname{u}=\ln{\left(x \right)}^{4}$$$ 和 $$$\operatorname{dv}=dx$$$。

则 $$$\operatorname{du}=\left(\ln{\left(x \right)}^{4}\right)^{\prime }dx=\frac{4 \ln{\left(x \right)}^{3}}{x} dx$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{1 d x}=x$$$ (步骤见 »)。

积分变为

$$\frac{{\color{red}{\int{\ln{\left(x \right)}^{4} d x}}}}{2}=\frac{{\color{red}{\left(\ln{\left(x \right)}^{4} \cdot x-\int{x \cdot \frac{4 \ln{\left(x \right)}^{3}}{x} d x}\right)}}}{2}=\frac{{\color{red}{\left(x \ln{\left(x \right)}^{4} - \int{4 \ln{\left(x \right)}^{3} d x}\right)}}}{2}$$

对 $$$c=4$$$ 和 $$$f{\left(x \right)} = \ln{\left(x \right)}^{3}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$\frac{x \ln{\left(x \right)}^{4}}{2} - \frac{{\color{red}{\int{4 \ln{\left(x \right)}^{3} d x}}}}{2} = \frac{x \ln{\left(x \right)}^{4}}{2} - \frac{{\color{red}{\left(4 \int{\ln{\left(x \right)}^{3} d x}\right)}}}{2}$$

对于积分$$$\int{\ln{\left(x \right)}^{3} d x}$$$，使用分部积分法$$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

设 $$$\operatorname{u}=\ln{\left(x \right)}^{3}$$$ 和 $$$\operatorname{dv}=dx$$$。

则 $$$\operatorname{du}=\left(\ln{\left(x \right)}^{3}\right)^{\prime }dx=\frac{3 \ln{\left(x \right)}^{2}}{x} dx$$$ (步骤见 »)，并且 $$$\operatorname{v}=\int{1 d x}=x$$$ (步骤见 »)。

该积分可以改写为

$$\frac{x \ln{\left(x \right)}^{4}}{2} - 2 {\color{red}{\int{\ln{\left(x \right)}^{3} d x}}}=\frac{x \ln{\left(x \right)}^{4}}{2} - 2 {\color{red}{\left(\ln{\left(x \right)}^{3} \cdot x-\int{x \cdot \frac{3 \ln{\left(x \right)}^{2}}{x} d x}\right)}}=\frac{x \ln{\left(x \right)}^{4}}{2} - 2 {\color{red}{\left(x \ln{\left(x \right)}^{3} - \int{3 \ln{\left(x \right)}^{2} d x}\right)}}$$

对 $$$c=3$$$ 和 $$$f{\left(x \right)} = \ln{\left(x \right)}^{2}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$\frac{x \ln{\left(x \right)}^{4}}{2} - 2 x \ln{\left(x \right)}^{3} + 2 {\color{red}{\int{3 \ln{\left(x \right)}^{2} d x}}} = \frac{x \ln{\left(x \right)}^{4}}{2} - 2 x \ln{\left(x \right)}^{3} + 2 {\color{red}{\left(3 \int{\ln{\left(x \right)}^{2} d x}\right)}}$$

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

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

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

所以，

$$\frac{x \ln{\left(x \right)}^{4}}{2} - 2 x \ln{\left(x \right)}^{3} + 6 {\color{red}{\int{\ln{\left(x \right)}^{2} d x}}}=\frac{x \ln{\left(x \right)}^{4}}{2} - 2 x \ln{\left(x \right)}^{3} + 6 {\color{red}{\left(\ln{\left(x \right)}^{2} \cdot x-\int{x \cdot \frac{2 \ln{\left(x \right)}}{x} d x}\right)}}=\frac{x \ln{\left(x \right)}^{4}}{2} - 2 x \ln{\left(x \right)}^{3} + 6 {\color{red}{\left(x \ln{\left(x \right)}^{2} - \int{2 \ln{\left(x \right)} d x}\right)}}$$

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

$$\frac{x \ln{\left(x \right)}^{4}}{2} - 2 x \ln{\left(x \right)}^{3} + 6 x \ln{\left(x \right)}^{2} - 6 {\color{red}{\int{2 \ln{\left(x \right)} d x}}} = \frac{x \ln{\left(x \right)}^{4}}{2} - 2 x \ln{\left(x \right)}^{3} + 6 x \ln{\left(x \right)}^{2} - 6 {\color{red}{\left(2 \int{\ln{\left(x \right)} d x}\right)}}$$

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

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

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

所以，

$$\frac{x \ln{\left(x \right)}^{4}}{2} - 2 x \ln{\left(x \right)}^{3} + 6 x \ln{\left(x \right)}^{2} - 12 {\color{red}{\int{\ln{\left(x \right)} d x}}}=\frac{x \ln{\left(x \right)}^{4}}{2} - 2 x \ln{\left(x \right)}^{3} + 6 x \ln{\left(x \right)}^{2} - 12 {\color{red}{\left(\ln{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x} d x}\right)}}=\frac{x \ln{\left(x \right)}^{4}}{2} - 2 x \ln{\left(x \right)}^{3} + 6 x \ln{\left(x \right)}^{2} - 12 {\color{red}{\left(x \ln{\left(x \right)} - \int{1 d x}\right)}}$$

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=1$$$：

$$\frac{x \ln{\left(x \right)}^{4}}{2} - 2 x \ln{\left(x \right)}^{3} + 6 x \ln{\left(x \right)}^{2} - 12 x \ln{\left(x \right)} + 12 {\color{red}{\int{1 d x}}} = \frac{x \ln{\left(x \right)}^{4}}{2} - 2 x \ln{\left(x \right)}^{3} + 6 x \ln{\left(x \right)}^{2} - 12 x \ln{\left(x \right)} + 12 {\color{red}{x}}$$

因此，

$$\int{\frac{\ln{\left(x \right)}^{4}}{2} d x} = \frac{x \ln{\left(x \right)}^{4}}{2} - 2 x \ln{\left(x \right)}^{3} + 6 x \ln{\left(x \right)}^{2} - 12 x \ln{\left(x \right)} + 12 x$$

化简：

$$\int{\frac{\ln{\left(x \right)}^{4}}{2} d x} = \frac{x \left(\ln{\left(x \right)}^{4} - 4 \ln{\left(x \right)}^{3} + 12 \ln{\left(x \right)}^{2} - 24 \ln{\left(x \right)} + 24\right)}{2}$$

加上积分常数：

$$\int{\frac{\ln{\left(x \right)}^{4}}{2} d x} = \frac{x \left(\ln{\left(x \right)}^{4} - 4 \ln{\left(x \right)}^{3} + 12 \ln{\left(x \right)}^{2} - 24 \ln{\left(x \right)} + 24\right)}{2}+C$$

$$$\int \frac{\ln^{4}\left(x\right)}{2}\, dx = \frac{x \left(\ln^{4}\left(x\right) - 4 \ln^{3}\left(x\right) + 12 \ln^{2}\left(x\right) - 24 \ln\left(x\right) + 24\right)}{2} + C$$$A