$$$\frac{6 \ln^{2}\left(x\right)}{x}$$$ 的积分

求$$$\int \frac{6 \ln^{2}\left(x\right)}{x}\, dx$$$。

对 $$$c=6$$$ 和 $$$f{\left(x \right)} = \frac{\ln{\left(x \right)}^{2}}{x}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

设$$$u=\ln{\left(x \right)}$$$。

则$$$du=\left(\ln{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x}$$$ (步骤见»)，并有$$$\frac{dx}{x} = du$$$。

该积分可以改写为

$$6 {\color{red}{\int{\frac{\ln{\left(x \right)}^{2}}{x} d x}}} = 6 {\color{red}{\int{u^{2} d u}}}$$

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

$$6 {\color{red}{\int{u^{2} d u}}}=6 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=6 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

回忆一下 $$$u=\ln{\left(x \right)}$$$:

$$2 {\color{red}{u}}^{3} = 2 {\color{red}{\ln{\left(x \right)}}}^{3}$$

因此，

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

加上积分常数：

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

$$$\int \frac{6 \ln^{2}\left(x\right)}{x}\, dx = 2 \ln^{3}\left(x\right) + C$$$A