$$$\frac{\ln^{3}\left(x\right)}{x^{2}}$$$ 的積分
您的輸入
求$$$\int \frac{\ln^{3}\left(x\right)}{x^{2}}\, dx$$$。
解答
令 $$$u=\frac{1}{x}$$$。
則 $$$du=\left(\frac{1}{x}\right)^{\prime }dx = - \frac{1}{x^{2}} dx$$$ (步驟見»),並可得 $$$\frac{dx}{x^{2}} = - du$$$。
該積分變為
$${\color{red}{\int{\frac{\ln{\left(x \right)}^{3}}{x^{2}} d x}}} = {\color{red}{\int{\ln{\left(u \right)}^{3} d u}}}$$
對於積分 $$$\int{\ln{\left(u \right)}^{3} d u}$$$,使用分部積分法 $$$\int \operatorname{\kappa} \operatorname{dv} = \operatorname{\kappa}\operatorname{v} - \int \operatorname{v} \operatorname{d\kappa}$$$。
令 $$$\operatorname{\kappa}=\ln{\left(u \right)}^{3}$$$ 與 $$$\operatorname{dv}=du$$$。
則 $$$\operatorname{d\kappa}=\left(\ln{\left(u \right)}^{3}\right)^{\prime }du=\frac{3 \ln{\left(u \right)}^{2}}{u} du$$$(步驟見 »),且 $$$\operatorname{v}=\int{1 d u}=u$$$(步驟見 »)。
該積分變為
$${\color{red}{\int{\ln{\left(u \right)}^{3} d u}}}={\color{red}{\left(\ln{\left(u \right)}^{3} \cdot u-\int{u \cdot \frac{3 \ln{\left(u \right)}^{2}}{u} d u}\right)}}={\color{red}{\left(u \ln{\left(u \right)}^{3} - \int{3 \ln{\left(u \right)}^{2} d u}\right)}}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=3$$$ 與 $$$f{\left(u \right)} = \ln{\left(u \right)}^{2}$$$:
$$u \ln{\left(u \right)}^{3} - {\color{red}{\int{3 \ln{\left(u \right)}^{2} d u}}} = u \ln{\left(u \right)}^{3} - {\color{red}{\left(3 \int{\ln{\left(u \right)}^{2} d u}\right)}}$$
對於積分 $$$\int{\ln{\left(u \right)}^{2} d u}$$$,使用分部積分法 $$$\int \operatorname{\kappa} \operatorname{dv} = \operatorname{\kappa}\operatorname{v} - \int \operatorname{v} \operatorname{d\kappa}$$$。
令 $$$\operatorname{\kappa}=\ln{\left(u \right)}^{2}$$$ 與 $$$\operatorname{dv}=du$$$。
則 $$$\operatorname{d\kappa}=\left(\ln{\left(u \right)}^{2}\right)^{\prime }du=\frac{2 \ln{\left(u \right)}}{u} du$$$(步驟見 »),且 $$$\operatorname{v}=\int{1 d u}=u$$$(步驟見 »)。
所以,
$$u \ln{\left(u \right)}^{3} - 3 {\color{red}{\int{\ln{\left(u \right)}^{2} d u}}}=u \ln{\left(u \right)}^{3} - 3 {\color{red}{\left(\ln{\left(u \right)}^{2} \cdot u-\int{u \cdot \frac{2 \ln{\left(u \right)}}{u} d u}\right)}}=u \ln{\left(u \right)}^{3} - 3 {\color{red}{\left(u \ln{\left(u \right)}^{2} - \int{2 \ln{\left(u \right)} d u}\right)}}$$
套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$,使用 $$$c=2$$$ 與 $$$f{\left(u \right)} = \ln{\left(u \right)}$$$:
$$u \ln{\left(u \right)}^{3} - 3 u \ln{\left(u \right)}^{2} + 3 {\color{red}{\int{2 \ln{\left(u \right)} d u}}} = u \ln{\left(u \right)}^{3} - 3 u \ln{\left(u \right)}^{2} + 3 {\color{red}{\left(2 \int{\ln{\left(u \right)} d u}\right)}}$$
對於積分 $$$\int{\ln{\left(u \right)} d u}$$$,使用分部積分法 $$$\int \operatorname{\kappa} \operatorname{dv} = \operatorname{\kappa}\operatorname{v} - \int \operatorname{v} \operatorname{d\kappa}$$$。
令 $$$\operatorname{\kappa}=\ln{\left(u \right)}$$$ 與 $$$\operatorname{dv}=du$$$。
則 $$$\operatorname{d\kappa}=\left(\ln{\left(u \right)}\right)^{\prime }du=\frac{du}{u}$$$(步驟見 »),且 $$$\operatorname{v}=\int{1 d u}=u$$$(步驟見 »)。
因此,
$$u \ln{\left(u \right)}^{3} - 3 u \ln{\left(u \right)}^{2} + 6 {\color{red}{\int{\ln{\left(u \right)} d u}}}=u \ln{\left(u \right)}^{3} - 3 u \ln{\left(u \right)}^{2} + 6 {\color{red}{\left(\ln{\left(u \right)} \cdot u-\int{u \cdot \frac{1}{u} d u}\right)}}=u \ln{\left(u \right)}^{3} - 3 u \ln{\left(u \right)}^{2} + 6 {\color{red}{\left(u \ln{\left(u \right)} - \int{1 d u}\right)}}$$
配合 $$$c=1$$$,應用常數法則 $$$\int c\, du = c u$$$:
$$u \ln{\left(u \right)}^{3} - 3 u \ln{\left(u \right)}^{2} + 6 u \ln{\left(u \right)} - 6 {\color{red}{\int{1 d u}}} = u \ln{\left(u \right)}^{3} - 3 u \ln{\left(u \right)}^{2} + 6 u \ln{\left(u \right)} - 6 {\color{red}{u}}$$
回顧一下 $$$u=\frac{1}{x}$$$:
$$- 6 {\color{red}{u}} + 6 {\color{red}{u}} \ln{\left({\color{red}{u}} \right)} - 3 {\color{red}{u}} \ln{\left({\color{red}{u}} \right)}^{2} + {\color{red}{u}} \ln{\left({\color{red}{u}} \right)}^{3} = - 6 {\color{red}{\frac{1}{x}}} + 6 {\color{red}{\frac{1}{x}}} \ln{\left({\color{red}{\frac{1}{x}}} \right)} - 3 {\color{red}{\frac{1}{x}}} \ln{\left({\color{red}{\frac{1}{x}}} \right)}^{2} + {\color{red}{\frac{1}{x}}} \ln{\left({\color{red}{\frac{1}{x}}} \right)}^{3}$$
因此,
$$\int{\frac{\ln{\left(x \right)}^{3}}{x^{2}} d x} = \frac{\ln{\left(\frac{1}{x} \right)}^{3}}{x} - \frac{3 \ln{\left(\frac{1}{x} \right)}^{2}}{x} + \frac{6 \ln{\left(\frac{1}{x} \right)}}{x} - \frac{6}{x}$$
化簡:
$$\int{\frac{\ln{\left(x \right)}^{3}}{x^{2}} d x} = \frac{- \ln{\left(x \right)}^{3} - 3 \ln{\left(x \right)}^{2} - 6 \ln{\left(x \right)} - 6}{x}$$
加上積分常數:
$$\int{\frac{\ln{\left(x \right)}^{3}}{x^{2}} d x} = \frac{- \ln{\left(x \right)}^{3} - 3 \ln{\left(x \right)}^{2} - 6 \ln{\left(x \right)} - 6}{x}+C$$
答案
$$$\int \frac{\ln^{3}\left(x\right)}{x^{2}}\, dx = \frac{- \ln^{3}\left(x\right) - 3 \ln^{2}\left(x\right) - 6 \ln\left(x\right) - 6}{x} + C$$$A