$$$\frac{x^{2} + 1}{x^{2} - 1}$$$ 的积分

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

由于分子次数不小于分母次数，进行多项式长除法（步骤见»）:

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

逐项积分：

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

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=1$$$：

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

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

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

进行部分分式分解（步骤可见»）:

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

逐项积分：

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

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

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

设$$$u=x - 1$$$。

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

因此，

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

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

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

回忆一下 $$$u=x - 1$$$:

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

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

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

设$$$u=x + 1$$$。

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

所以，

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

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

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

回忆一下 $$$u=x + 1$$$:

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

因此，

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

加上积分常数：

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

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