$$$\frac{\ln\left(1 - x^{2}\right)}{2}$$$ 的積分

求$$$\int \frac{\ln\left(1 - x^{2}\right)}{2}\, dx$$$。

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 $$$f{\left(x \right)} = \ln{\left(1 - x^{2} \right)}$$$：

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

對於積分 $$$\int{\ln{\left(1 - x^{2} \right)} d x}$$$，使用分部積分法 $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$。

令 $$$\operatorname{u}=\ln{\left(1 - x^{2} \right)}$$$ 與 $$$\operatorname{dv}=dx$$$。

則 $$$\operatorname{du}=\left(\ln{\left(1 - x^{2} \right)}\right)^{\prime }dx=\frac{2 x}{x^{2} - 1} dx$$$（步驟見 »），且 $$$\operatorname{v}=\int{1 d x}=x$$$（步驟見 »）。

該積分變為

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

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

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

由於分子次數不小於分母次數，進行多項式長除法（步驟見»）:

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

逐項積分：

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

配合 $$$c=1$$$，應用常數法則 $$$\int c\, dx = c x$$$：

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

進行部分分式分解（步驟可見 »）:

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

逐項積分：

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

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 $$$f{\left(x \right)} = \frac{1}{x - 1}$$$：

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

令 $$$u=x - 1$$$。

則 $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

因此，

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

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

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

回顧一下 $$$u=x - 1$$$：

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

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

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

令 $$$u=x + 1$$$。

則 $$$du=\left(x + 1\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

該積分變為

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

$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：

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

回顧一下 $$$u=x + 1$$$：

$$\frac{x \ln{\left(1 - x^{2} \right)}}{2} - x - \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = \frac{x \ln{\left(1 - x^{2} \right)}}{2} - x - \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{{\color{red}{\left(x + 1\right)}}}\right| \right)}}{2}$$

因此，

$$\int{\frac{\ln{\left(1 - x^{2} \right)}}{2} d x} = \frac{x \ln{\left(1 - x^{2} \right)}}{2} - x - \frac{\ln{\left(\left|{x - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{x + 1}\right| \right)}}{2}$$

加上積分常數：

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

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