$$$\frac{1}{x^{2} - 64}$$$ 的積分

求$$$\int \frac{1}{x^{2} - 64}\, dx$$$。

進行部分分式分解（步驟可見 »）:

$${\color{red}{\int{\frac{1}{x^{2} - 64} d x}}} = {\color{red}{\int{\left(- \frac{1}{16 \left(x + 8\right)} + \frac{1}{16 \left(x - 8\right)}\right)d x}}}$$

逐項積分：

$${\color{red}{\int{\left(- \frac{1}{16 \left(x + 8\right)} + \frac{1}{16 \left(x - 8\right)}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{16 \left(x - 8\right)} d x} - \int{\frac{1}{16 \left(x + 8\right)} d x}\right)}}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{16}$$$ 與 $$$f{\left(x \right)} = \frac{1}{x + 8}$$$：

$$\int{\frac{1}{16 \left(x - 8\right)} d x} - {\color{red}{\int{\frac{1}{16 \left(x + 8\right)} d x}}} = \int{\frac{1}{16 \left(x - 8\right)} d x} - {\color{red}{\left(\frac{\int{\frac{1}{x + 8} d x}}{16}\right)}}$$

令 $$$u=x + 8$$$。

則 $$$du=\left(x + 8\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

因此，

$$\int{\frac{1}{16 \left(x - 8\right)} d x} - \frac{{\color{red}{\int{\frac{1}{x + 8} d x}}}}{16} = \int{\frac{1}{16 \left(x - 8\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{16}$$

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

$$\int{\frac{1}{16 \left(x - 8\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{16} = \int{\frac{1}{16 \left(x - 8\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{16}$$

回顧一下 $$$u=x + 8$$$：

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{16} + \int{\frac{1}{16 \left(x - 8\right)} d x} = - \frac{\ln{\left(\left|{{\color{red}{\left(x + 8\right)}}}\right| \right)}}{16} + \int{\frac{1}{16 \left(x - 8\right)} d x}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{16}$$$ 與 $$$f{\left(x \right)} = \frac{1}{x - 8}$$$：

$$- \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + {\color{red}{\int{\frac{1}{16 \left(x - 8\right)} d x}}} = - \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + {\color{red}{\left(\frac{\int{\frac{1}{x - 8} d x}}{16}\right)}}$$

令 $$$u=x - 8$$$。

則 $$$du=\left(x - 8\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

該積分變為

$$- \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + \frac{{\color{red}{\int{\frac{1}{x - 8} d x}}}}{16} = - \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{16}$$

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

$$- \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{16} = - \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{16}$$

回顧一下 $$$u=x - 8$$$：

$$- \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{16} = - \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16} + \frac{\ln{\left(\left|{{\color{red}{\left(x - 8\right)}}}\right| \right)}}{16}$$

因此，

$$\int{\frac{1}{x^{2} - 64} d x} = \frac{\ln{\left(\left|{x - 8}\right| \right)}}{16} - \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16}$$

加上積分常數：

$$\int{\frac{1}{x^{2} - 64} d x} = \frac{\ln{\left(\left|{x - 8}\right| \right)}}{16} - \frac{\ln{\left(\left|{x + 8}\right| \right)}}{16}+C$$

$$$\int \frac{1}{x^{2} - 64}\, dx = \left(\frac{\ln\left(\left|{x - 8}\right|\right)}{16} - \frac{\ln\left(\left|{x + 8}\right|\right)}{16}\right) + C$$$A