$$$\frac{\ln^{2}\left(x^{2}\right)}{x}$$$ 的積分

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

已將輸入重寫為：$$$\int{\frac{\ln{\left(x^{2} \right)}^{2}}{x} d x}=\int{\frac{4 \ln{\left(x \right)}^{2}}{x} d x}$$$。

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=4$$$ 與 $$$f{\left(x \right)} = \frac{\ln{\left(x \right)}^{2}}{x}$$$：

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

令 $$$u=\ln{\left(x \right)}$$$。

則 $$$du=\left(\ln{\left(x \right)}\right)^{\prime }dx = \frac{dx}{x}$$$ (步驟見»)，並可得 $$$\frac{dx}{x} = du$$$。

因此，

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

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=2$$$：

$$4 {\color{red}{\int{u^{2} d u}}}=4 {\color{red}{\frac{u^{1 + 2}}{1 + 2}}}=4 {\color{red}{\left(\frac{u^{3}}{3}\right)}}$$

回顧一下 $$$u=\ln{\left(x \right)}$$$：

$$\frac{4 {\color{red}{u}}^{3}}{3} = \frac{4 {\color{red}{\ln{\left(x \right)}}}^{3}}{3}$$

因此，

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

加上積分常數：

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

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