$$$\ln\left(1 - x^{2}\right)$$$ 的積分

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

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

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

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

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

由於分子次數不小於分母次數，進行多項式長除法（步驟見»）:

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

逐項積分：

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

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

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

進行部分分式分解（步驟可見 »）:

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

逐項積分：

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

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

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

令 $$$u=x - 1$$$。

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

該積分可改寫為

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

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

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

回顧一下 $$$u=x - 1$$$：

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

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

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

令 $$$u=x + 1$$$。

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

因此，

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

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

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

回顧一下 $$$u=x + 1$$$：

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

因此，

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

加上積分常數：

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

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