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

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

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

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

進行部分分式分解（步驟可見 »）:

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

逐項積分：

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

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

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

令 $$$u=x + 1$$$。

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

所以，

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

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

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

回顧一下 $$$u=x + 1$$$：

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

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

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

令 $$$u=x - 1$$$。

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

因此，

$$\frac{\ln{\left(\left|{x + 1}\right| \right)}}{4} - \frac{{\color{red}{\int{\frac{1}{x - 1} d x}}}}{4} = \frac{\ln{\left(\left|{x + 1}\right| \right)}}{4} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{4}$$

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

$$\frac{\ln{\left(\left|{x + 1}\right| \right)}}{4} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{4} = \frac{\ln{\left(\left|{x + 1}\right| \right)}}{4} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{4}$$

回顧一下 $$$u=x - 1$$$：

$$\frac{\ln{\left(\left|{x + 1}\right| \right)}}{4} - \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{4} = \frac{\ln{\left(\left|{x + 1}\right| \right)}}{4} - \frac{\ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}}{4}$$

因此，

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

化簡：

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

加上積分常數：

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

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