$$$x^{3} \ln^{2}\left(x\right)$$$ 的積分

求$$$\int x^{3} \ln^{2}\left(x\right)\, dx$$$。

對於積分 $$$\int{x^{3} \ln{\left(x \right)}^{2} d x}$$$，使用分部積分法 $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

令 $$$\operatorname{u}=\ln{\left(x \right)}^{2}$$$ 與 $$$\operatorname{dv}=x^{3} dx$$$。

則 $$$\operatorname{du}=\left(\ln{\left(x \right)}^{2}\right)^{\prime }dx=\frac{2 \ln{\left(x \right)}}{x} dx$$$（步驟見 »），且 $$$\operatorname{v}=\int{x^{3} d x}=\frac{x^{4}}{4}$$$（步驟見 »）。

該積分變為

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(x \right)} = x^{3} \ln{\left(x \right)}$$$：

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

對於積分 $$$\int{x^{3} \ln{\left(x \right)} d x}$$$，使用分部積分法 $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

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

則 $$$\operatorname{du}=\left(\ln{\left(x \right)}\right)^{\prime }dx=\frac{dx}{x}$$$（步驟見 »），且 $$$\operatorname{v}=\int{x^{3} d x}=\frac{x^{4}}{4}$$$（步驟見 »）。

所以，

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

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{4}$$$ 與 $$$f{\left(x \right)} = x^{3}$$$：

$$\frac{x^{4} \ln{\left(x \right)}^{2}}{4} - \frac{x^{4} \ln{\left(x \right)}}{8} + \frac{{\color{red}{\int{\frac{x^{3}}{4} d x}}}}{2} = \frac{x^{4} \ln{\left(x \right)}^{2}}{4} - \frac{x^{4} \ln{\left(x \right)}}{8} + \frac{{\color{red}{\left(\frac{\int{x^{3} d x}}{4}\right)}}}{2}$$

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=3$$$：

$$\frac{x^{4} \ln{\left(x \right)}^{2}}{4} - \frac{x^{4} \ln{\left(x \right)}}{8} + \frac{{\color{red}{\int{x^{3} d x}}}}{8}=\frac{x^{4} \ln{\left(x \right)}^{2}}{4} - \frac{x^{4} \ln{\left(x \right)}}{8} + \frac{{\color{red}{\frac{x^{1 + 3}}{1 + 3}}}}{8}=\frac{x^{4} \ln{\left(x \right)}^{2}}{4} - \frac{x^{4} \ln{\left(x \right)}}{8} + \frac{{\color{red}{\left(\frac{x^{4}}{4}\right)}}}{8}$$

因此，

$$\int{x^{3} \ln{\left(x \right)}^{2} d x} = \frac{x^{4} \ln{\left(x \right)}^{2}}{4} - \frac{x^{4} \ln{\left(x \right)}}{8} + \frac{x^{4}}{32}$$

化簡：

$$\int{x^{3} \ln{\left(x \right)}^{2} d x} = \frac{x^{4} \left(8 \ln{\left(x \right)}^{2} - 4 \ln{\left(x \right)} + 1\right)}{32}$$

加上積分常數：

$$\int{x^{3} \ln{\left(x \right)}^{2} d x} = \frac{x^{4} \left(8 \ln{\left(x \right)}^{2} - 4 \ln{\left(x \right)} + 1\right)}{32}+C$$

$$$\int x^{3} \ln^{2}\left(x\right)\, dx = \frac{x^{4} \left(8 \ln^{2}\left(x\right) - 4 \ln\left(x\right) + 1\right)}{32} + C$$$A