$$$\frac{1}{- a^{2} + x^{2}}$$$ 关于$$$x$$$的积分

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

进行部分分式分解:

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

逐项积分：

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

对 $$$c=\frac{1}{2 \left|{a}\right|}$$$ 和 $$$f{\left(x \right)} = \frac{1}{- a + x}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

设$$$u=- a + x$$$。

则$$$du=\left(- a + x\right)^{\prime }dx = 1 dx$$$ (步骤见»)，并有$$$dx = du$$$。

因此，

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

$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

回忆一下 $$$u=- a + x$$$:

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

对 $$$c=\frac{1}{2 \left|{a}\right|}$$$ 和 $$$f{\left(x \right)} = \frac{1}{a + x}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

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

设$$$u=a + x$$$。

则$$$du=\left(a + x\right)^{\prime }dx = 1 dx$$$ (步骤见»)，并有$$$dx = du$$$。

因此，

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

$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

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

回忆一下 $$$u=a + x$$$:

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

因此，

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

化简：

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

加上积分常数：

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

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