$$$\frac{1}{x^{2} - 16}$$$ 的積分

求$$$\int \frac{1}{x^{2} - 16}\, dx$$$。

進行部分分式分解（步驟可見 »）:

$${\color{red}{\int{\frac{1}{x^{2} - 16} d x}}} = {\color{red}{\int{\left(- \frac{1}{8 \left(x + 4\right)} + \frac{1}{8 \left(x - 4\right)}\right)d x}}}$$

逐項積分：

$${\color{red}{\int{\left(- \frac{1}{8 \left(x + 4\right)} + \frac{1}{8 \left(x - 4\right)}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{8 \left(x - 4\right)} d x} - \int{\frac{1}{8 \left(x + 4\right)} d x}\right)}}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{8}$$$ 與 $$$f{\left(x \right)} = \frac{1}{x + 4}$$$：

$$\int{\frac{1}{8 \left(x - 4\right)} d x} - {\color{red}{\int{\frac{1}{8 \left(x + 4\right)} d x}}} = \int{\frac{1}{8 \left(x - 4\right)} d x} - {\color{red}{\left(\frac{\int{\frac{1}{x + 4} d x}}{8}\right)}}$$

令 $$$u=x + 4$$$。

則 $$$du=\left(x + 4\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

因此，

$$\int{\frac{1}{8 \left(x - 4\right)} d x} - \frac{{\color{red}{\int{\frac{1}{x + 4} d x}}}}{8} = \int{\frac{1}{8 \left(x - 4\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{8}$$

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

$$\int{\frac{1}{8 \left(x - 4\right)} d x} - \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{8} = \int{\frac{1}{8 \left(x - 4\right)} d x} - \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}$$

回顧一下 $$$u=x + 4$$$：

$$- \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} + \int{\frac{1}{8 \left(x - 4\right)} d x} = - \frac{\ln{\left(\left|{{\color{red}{\left(x + 4\right)}}}\right| \right)}}{8} + \int{\frac{1}{8 \left(x - 4\right)} d x}$$

套用常數倍法則 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$，使用 $$$c=\frac{1}{8}$$$ 與 $$$f{\left(x \right)} = \frac{1}{x - 4}$$$：

$$- \frac{\ln{\left(\left|{x + 4}\right| \right)}}{8} + {\color{red}{\int{\frac{1}{8 \left(x - 4\right)} d x}}} = - \frac{\ln{\left(\left|{x + 4}\right| \right)}}{8} + {\color{red}{\left(\frac{\int{\frac{1}{x - 4} d x}}{8}\right)}}$$

令 $$$u=x - 4$$$。

則 $$$du=\left(x - 4\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

所以，

$$- \frac{\ln{\left(\left|{x + 4}\right| \right)}}{8} + \frac{{\color{red}{\int{\frac{1}{x - 4} d x}}}}{8} = - \frac{\ln{\left(\left|{x + 4}\right| \right)}}{8} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{8}$$

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

$$- \frac{\ln{\left(\left|{x + 4}\right| \right)}}{8} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{8} = - \frac{\ln{\left(\left|{x + 4}\right| \right)}}{8} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{8}$$

回顧一下 $$$u=x - 4$$$：

$$- \frac{\ln{\left(\left|{x + 4}\right| \right)}}{8} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{8} = - \frac{\ln{\left(\left|{x + 4}\right| \right)}}{8} + \frac{\ln{\left(\left|{{\color{red}{\left(x - 4\right)}}}\right| \right)}}{8}$$

因此，

$$\int{\frac{1}{x^{2} - 16} d x} = \frac{\ln{\left(\left|{x - 4}\right| \right)}}{8} - \frac{\ln{\left(\left|{x + 4}\right| \right)}}{8}$$

加上積分常數：

$$\int{\frac{1}{x^{2} - 16} d x} = \frac{\ln{\left(\left|{x - 4}\right| \right)}}{8} - \frac{\ln{\left(\left|{x + 4}\right| \right)}}{8}+C$$

$$$\int \frac{1}{x^{2} - 16}\, dx = \left(\frac{\ln\left(\left|{x - 4}\right|\right)}{8} - \frac{\ln\left(\left|{x + 4}\right|\right)}{8}\right) + C$$$A