$$$\frac{1}{1 - t^{2}}$$$ 的积分

求$$$\int \frac{1}{1 - t^{2}}\, dt$$$。

进行部分分式分解（步骤可见»）:

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

逐项积分：

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

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

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

设$$$u=t + 1$$$。

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

所以，

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

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

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

回忆一下 $$$u=t + 1$$$:

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

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

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

设$$$u=t - 1$$$。

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

所以，

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

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

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

回忆一下 $$$u=t - 1$$$:

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

因此，

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

化简：

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

加上积分常数：

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

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