$$$\frac{t^{2}}{t^{2} - 2}$$$ 的积分

求$$$\int \frac{t^{2}}{t^{2} - 2}\, dt$$$。

改写并拆分该分式:

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

逐项积分：

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

应用常数法则 $$$\int c\, dt = c t$$$，使用 $$$c=1$$$：

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

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

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

进行部分分式分解（步骤可见»）:

$$t + 2 {\color{red}{\int{\frac{1}{t^{2} - 2} d t}}} = t + 2 {\color{red}{\int{\left(- \frac{\sqrt{2}}{4 \left(t + \sqrt{2}\right)} + \frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)}\right)d t}}}$$

逐项积分：

$$t + 2 {\color{red}{\int{\left(- \frac{\sqrt{2}}{4 \left(t + \sqrt{2}\right)} + \frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)}\right)d t}}} = t + 2 {\color{red}{\left(\int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - \int{\frac{\sqrt{2}}{4 \left(t + \sqrt{2}\right)} d t}\right)}}$$

对 $$$c=\frac{\sqrt{2}}{4}$$$ 和 $$$f{\left(t \right)} = \frac{1}{t + \sqrt{2}}$$$ 应用常数倍法则 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$：

$$t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - 2 {\color{red}{\int{\frac{\sqrt{2}}{4 \left(t + \sqrt{2}\right)} d t}}} = t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - 2 {\color{red}{\left(\frac{\sqrt{2} \int{\frac{1}{t + \sqrt{2}} d t}}{4}\right)}}$$

设$$$u=t + \sqrt{2}$$$。

则$$$du=\left(t + \sqrt{2}\right)^{\prime }dt = 1 dt$$$ (步骤见»)，并有$$$dt = du$$$。

因此，

$$t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - \frac{\sqrt{2} {\color{red}{\int{\frac{1}{t + \sqrt{2}} d t}}}}{2} = t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - \frac{\sqrt{2} {\color{red}{\int{\frac{1}{u} d u}}}}{2}$$

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

$$t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - \frac{\sqrt{2} {\color{red}{\int{\frac{1}{u} d u}}}}{2} = t + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} - \frac{\sqrt{2} {\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$

回忆一下 $$$u=t + \sqrt{2}$$$:

$$t - \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t} = t - \frac{\sqrt{2} \ln{\left(\left|{{\color{red}{\left(t + \sqrt{2}\right)}}}\right| \right)}}{2} + 2 \int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t}$$

对 $$$c=\frac{\sqrt{2}}{4}$$$ 和 $$$f{\left(t \right)} = \frac{1}{t - \sqrt{2}}$$$ 应用常数倍法则 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$：

$$t - \frac{\sqrt{2} \ln{\left(\left|{t + \sqrt{2}}\right| \right)}}{2} + 2 {\color{red}{\int{\frac{\sqrt{2}}{4 \left(t - \sqrt{2}\right)} d t}}} = t - \frac{\sqrt{2} \ln{\left(\left|{t + \sqrt{2}}\right| \right)}}{2} + 2 {\color{red}{\left(\frac{\sqrt{2} \int{\frac{1}{t - \sqrt{2}} d t}}{4}\right)}}$$

设$$$u=t - \sqrt{2}$$$。

则$$$du=\left(t - \sqrt{2}\right)^{\prime }dt = 1 dt$$$ (步骤见»)，并有$$$dt = du$$$。

所以，

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

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

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

回忆一下 $$$u=t - \sqrt{2}$$$:

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

因此，

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

加上积分常数：

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

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