$$$\frac{5}{1 - x^{2}}$$$ 的积分

求$$$\int \frac{5}{1 - x^{2}}\, dx$$$。

对 $$$c=5$$$ 和 $$$f{\left(x \right)} = \frac{1}{1 - x^{2}}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\frac{5}{1 - x^{2}} d x}}} = {\color{red}{\left(5 \int{\frac{1}{1 - x^{2}} d x}\right)}}$$

进行部分分式分解（步骤可见»）:

$$5 {\color{red}{\int{\frac{1}{1 - x^{2}} d x}}} = 5 {\color{red}{\int{\left(\frac{1}{2 \left(x + 1\right)} - \frac{1}{2 \left(x - 1\right)}\right)d x}}}$$

逐项积分：

$$5 {\color{red}{\int{\left(\frac{1}{2 \left(x + 1\right)} - \frac{1}{2 \left(x - 1\right)}\right)d x}}} = 5 {\color{red}{\left(- \int{\frac{1}{2 \left(x - 1\right)} d x} + \int{\frac{1}{2 \left(x + 1\right)} d x}\right)}}$$

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

$$- 5 \int{\frac{1}{2 \left(x - 1\right)} d x} + 5 {\color{red}{\int{\frac{1}{2 \left(x + 1\right)} d x}}} = - 5 \int{\frac{1}{2 \left(x - 1\right)} d x} + 5 {\color{red}{\left(\frac{\int{\frac{1}{x + 1} d x}}{2}\right)}}$$

设$$$u=x + 1$$$。

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

因此，

$$- 5 \int{\frac{1}{2 \left(x - 1\right)} d x} + \frac{5 {\color{red}{\int{\frac{1}{x + 1} d x}}}}{2} = - 5 \int{\frac{1}{2 \left(x - 1\right)} d x} + \frac{5 {\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

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

$$- 5 \int{\frac{1}{2 \left(x - 1\right)} d x} + \frac{5 {\color{red}{\int{\frac{1}{u} d u}}}}{2} = - 5 \int{\frac{1}{2 \left(x - 1\right)} d x} + \frac{5 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

回忆一下 $$$u=x + 1$$$:

$$\frac{5 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} - 5 \int{\frac{1}{2 \left(x - 1\right)} d x} = \frac{5 \ln{\left(\left|{{\color{red}{\left(x + 1\right)}}}\right| \right)}}{2} - 5 \int{\frac{1}{2 \left(x - 1\right)} d x}$$

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

$$\frac{5 \ln{\left(\left|{x + 1}\right| \right)}}{2} - 5 {\color{red}{\int{\frac{1}{2 \left(x - 1\right)} d x}}} = \frac{5 \ln{\left(\left|{x + 1}\right| \right)}}{2} - 5 {\color{red}{\left(\frac{\int{\frac{1}{x - 1} d x}}{2}\right)}}$$

设$$$u=x - 1$$$。

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

该积分可以改写为

$$\frac{5 \ln{\left(\left|{x + 1}\right| \right)}}{2} - \frac{5 {\color{red}{\int{\frac{1}{x - 1} d x}}}}{2} = \frac{5 \ln{\left(\left|{x + 1}\right| \right)}}{2} - \frac{5 {\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

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

$$\frac{5 \ln{\left(\left|{x + 1}\right| \right)}}{2} - \frac{5 {\color{red}{\int{\frac{1}{u} d u}}}}{2} = \frac{5 \ln{\left(\left|{x + 1}\right| \right)}}{2} - \frac{5 {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

回忆一下 $$$u=x - 1$$$:

$$\frac{5 \ln{\left(\left|{x + 1}\right| \right)}}{2} - \frac{5 \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} = \frac{5 \ln{\left(\left|{x + 1}\right| \right)}}{2} - \frac{5 \ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}}{2}$$

因此，

$$\int{\frac{5}{1 - x^{2}} d x} = - \frac{5 \ln{\left(\left|{x - 1}\right| \right)}}{2} + \frac{5 \ln{\left(\left|{x + 1}\right| \right)}}{2}$$

化简：

$$\int{\frac{5}{1 - x^{2}} d x} = \frac{5 \left(- \ln{\left(\left|{x - 1}\right| \right)} + \ln{\left(\left|{x + 1}\right| \right)}\right)}{2}$$

加上积分常数：

$$\int{\frac{5}{1 - x^{2}} d x} = \frac{5 \left(- \ln{\left(\left|{x - 1}\right| \right)} + \ln{\left(\left|{x + 1}\right| \right)}\right)}{2}+C$$

$$$\int \frac{5}{1 - x^{2}}\, dx = \frac{5 \left(- \ln\left(\left|{x - 1}\right|\right) + \ln\left(\left|{x + 1}\right|\right)\right)}{2} + C$$$A