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$$$\frac{1}{t^{3} - t}$$$ 的积分
您的输入
求$$$\int \frac{1}{t^{3} - t}\, dt$$$。
解答
进行部分分式分解（步骤可见»）:
$${\color{red}{\int{\frac{1}{t^{3} - t} d t}}} = {\color{red}{\int{\left(\frac{1}{2 \left(t + 1\right)} + \frac{1}{2 \left(t - 1\right)} - \frac{1}{t}\right)d t}}}$$
逐项积分：
$${\color{red}{\int{\left(\frac{1}{2 \left(t + 1\right)} + \frac{1}{2 \left(t - 1\right)} - \frac{1}{t}\right)d t}}} = {\color{red}{\left(- \int{\frac{1}{t} d t} + \int{\frac{1}{2 \left(t - 1\right)} d t} + \int{\frac{1}{2 \left(t + 1\right)} d t}\right)}}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(t \right)} = \frac{1}{t + 1}$$$ 应用常数倍法则 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$：
$$- \int{\frac{1}{t} d t} + \int{\frac{1}{2 \left(t - 1\right)} d t} + {\color{red}{\int{\frac{1}{2 \left(t + 1\right)} d t}}} = - \int{\frac{1}{t} d t} + \int{\frac{1}{2 \left(t - 1\right)} d t} + {\color{red}{\left(\frac{\int{\frac{1}{t + 1} d t}}{2}\right)}}$$
设$$$u=t + 1$$$。
则$$$du=\left(t + 1\right)^{\prime }dt = 1 dt$$$ (步骤见»)，并有$$$dt = du$$$。
因此，
$$- \int{\frac{1}{t} d t} + \int{\frac{1}{2 \left(t - 1\right)} d t} + \frac{{\color{red}{\int{\frac{1}{t + 1} d t}}}}{2} = - \int{\frac{1}{t} d t} + \int{\frac{1}{2 \left(t - 1\right)} d t} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}$$
$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$- \int{\frac{1}{t} d t} + \int{\frac{1}{2 \left(t - 1\right)} d t} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = - \int{\frac{1}{t} d t} + \int{\frac{1}{2 \left(t - 1\right)} d t} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$
回忆一下 $$$u=t + 1$$$:
$$\frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} - \int{\frac{1}{t} d t} + \int{\frac{1}{2 \left(t - 1\right)} d t} = \frac{\ln{\left(\left|{{\color{red}{\left(t + 1\right)}}}\right| \right)}}{2} - \int{\frac{1}{t} d t} + \int{\frac{1}{2 \left(t - 1\right)} d t}$$
对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(t \right)} = \frac{1}{t - 1}$$$ 应用常数倍法则 $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$：
$$\frac{\ln{\left(\left|{t + 1}\right| \right)}}{2} - \int{\frac{1}{t} d t} + {\color{red}{\int{\frac{1}{2 \left(t - 1\right)} d t}}} = \frac{\ln{\left(\left|{t + 1}\right| \right)}}{2} - \int{\frac{1}{t} d t} + {\color{red}{\left(\frac{\int{\frac{1}{t - 1} d t}}{2}\right)}}$$
设$$$u=t - 1$$$。
则$$$du=\left(t - 1\right)^{\prime }dt = 1 dt$$$ (步骤见»)，并有$$$dt = du$$$。
因此，
$$\frac{\ln{\left(\left|{t + 1}\right| \right)}}{2} - \int{\frac{1}{t} d t} + \frac{{\color{red}{\int{\frac{1}{t - 1} d t}}}}{2} = \frac{\ln{\left(\left|{t + 1}\right| \right)}}{2} - \int{\frac{1}{t} d t} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2}$$
$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$\frac{\ln{\left(\left|{t + 1}\right| \right)}}{2} - \int{\frac{1}{t} d t} + \frac{{\color{red}{\int{\frac{1}{u} d u}}}}{2} = \frac{\ln{\left(\left|{t + 1}\right| \right)}}{2} - \int{\frac{1}{t} d t} + \frac{{\color{red}{\ln{\left(\left|{u}\right| \right)}}}}{2}$$
回忆一下 $$$u=t - 1$$$:
$$\frac{\ln{\left(\left|{t + 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{{\color{red}{u}}}\right| \right)}}{2} - \int{\frac{1}{t} d t} = \frac{\ln{\left(\left|{t + 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{{\color{red}{\left(t - 1\right)}}}\right| \right)}}{2} - \int{\frac{1}{t} d t}$$
$$$\frac{1}{t}$$$ 的积分为 $$$\int{\frac{1}{t} d t} = \ln{\left(\left|{t}\right| \right)}$$$:
$$\frac{\ln{\left(\left|{t - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{t + 1}\right| \right)}}{2} - {\color{red}{\int{\frac{1}{t} d t}}} = \frac{\ln{\left(\left|{t - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{t + 1}\right| \right)}}{2} - {\color{red}{\ln{\left(\left|{t}\right| \right)}}}$$
因此，
$$\int{\frac{1}{t^{3} - t} d t} = - \ln{\left(\left|{t}\right| \right)} + \frac{\ln{\left(\left|{t - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{t + 1}\right| \right)}}{2}$$
加上积分常数：
$$\int{\frac{1}{t^{3} - t} d t} = - \ln{\left(\left|{t}\right| \right)} + \frac{\ln{\left(\left|{t - 1}\right| \right)}}{2} + \frac{\ln{\left(\left|{t + 1}\right| \right)}}{2}+C$$
答案
$$$\int \frac{1}{t^{3} - t}\, dt = \left(- \ln\left(\left|{t}\right|\right) + \frac{\ln\left(\left|{t - 1}\right|\right)}{2} + \frac{\ln\left(\left|{t + 1}\right|\right)}{2}\right) + C$$$A