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

求$$$\int \frac{y}{x^{2} - 1}\, dx$$$。

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

$${\color{red}{\int{\frac{y}{x^{2} - 1} d x}}} = {\color{red}{y \int{\frac{1}{x^{2} - 1} d x}}}$$

进行部分分式分解（步骤可见»）:

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

逐项积分：

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

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

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

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

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

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

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

回忆一下 $$$u=x - 1$$$:

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

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

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

设$$$u=x + 1$$$。

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

因此，

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

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

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

回忆一下 $$$u=x + 1$$$:

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

因此，

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

化简：

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

加上积分常数：

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

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