$$$\frac{1}{- a^{2} + x^{2}}$$$ 對 $$$x$$$ 的積分

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

進行部分分式分解:

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

逐項積分：

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

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

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

令 $$$u=- a + x$$$。

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

因此，

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

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

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

回顧一下 $$$u=- a + x$$$：

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

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

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

令 $$$u=a + x$$$。

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

該積分可改寫為

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

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

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

回顧一下 $$$u=a + x$$$：

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

因此，

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

化簡：

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

加上積分常數：

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

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