$$$\operatorname{atanh}{\left(x \right)}$$$ 的積分

求$$$\int \operatorname{atanh}{\left(x \right)}\, dx$$$。

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

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

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

該積分可改寫為

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

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

$$x \operatorname{atanh}{\left(x \right)} - {\color{red}{\int{\left(- \frac{x}{\left(x - 1\right) \left(x + 1\right)}\right)d x}}} = x \operatorname{atanh}{\left(x \right)} - {\color{red}{\left(- \int{\frac{x}{\left(x - 1\right) \left(x + 1\right)} d x}\right)}}$$

進行部分分式分解（步驟可見 »）:

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

逐項積分：

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

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

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

令 $$$u=x + 1$$$。

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

該積分可改寫為

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

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

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

回顧一下 $$$u=x + 1$$$：

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

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

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

令 $$$u=x - 1$$$。

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

因此，

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

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

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

回顧一下 $$$u=x - 1$$$：

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

因此，

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

加上積分常數：

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

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