$$$\ln^{2}\left(x^{2}\right)$$$ 的积分

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

输入已重写为：$$$\int{\ln{\left(x^{2} \right)}^{2} d x}=\int{4 \ln{\left(x \right)}^{2} d x}$$$。

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

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

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 »)。

该积分可以改写为

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

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

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

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 »)。

因此，

$$4 x \ln{\left(x \right)}^{2} - 8 {\color{red}{\int{\ln{\left(x \right)} d x}}}=4 x \ln{\left(x \right)}^{2} - 8 {\color{red}{\left(\ln{\left(x \right)} \cdot x-\int{x \cdot \frac{1}{x} d x}\right)}}=4 x \ln{\left(x \right)}^{2} - 8 {\color{red}{\left(x \ln{\left(x \right)} - \int{1 d x}\right)}}$$

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

$$4 x \ln{\left(x \right)}^{2} - 8 x \ln{\left(x \right)} + 8 {\color{red}{\int{1 d x}}} = 4 x \ln{\left(x \right)}^{2} - 8 x \ln{\left(x \right)} + 8 {\color{red}{x}}$$

因此，

$$\int{4 \ln{\left(x \right)}^{2} d x} = 4 x \ln{\left(x \right)}^{2} - 8 x \ln{\left(x \right)} + 8 x$$

化简：

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

加上积分常数：

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

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