$$$\frac{1}{\left(x - 2\right) \left(x - 1\right)}$$$ 的積分
您的輸入
求$$$\int \frac{1}{\left(x - 2\right) \left(x - 1\right)}\, dx$$$。
解答
進行部分分式分解（步驟可見 »）:
$${\color{red}{\int{\frac{1}{\left(x - 2\right) \left(x - 1\right)} d x}}} = {\color{red}{\int{\left(- \frac{1}{x - 1} + \frac{1}{x - 2}\right)d x}}}$$
逐項積分：
$${\color{red}{\int{\left(- \frac{1}{x - 1} + \frac{1}{x - 2}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{x - 2} d x} - \int{\frac{1}{x - 1} d x}\right)}}$$
令 $$$u=x - 2$$$。
則 $$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。
該積分可改寫為
$$- \int{\frac{1}{x - 1} d x} + {\color{red}{\int{\frac{1}{x - 2} d x}}} = - \int{\frac{1}{x - 1} d x} + {\color{red}{\int{\frac{1}{u} d u}}}$$
$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：
$$- \int{\frac{1}{x - 1} d x} + {\color{red}{\int{\frac{1}{u} d u}}} = - \int{\frac{1}{x - 1} d x} + {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
回顧一下 $$$u=x - 2$$$：
$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} - \int{\frac{1}{x - 1} d x} = \ln{\left(\left|{{\color{red}{\left(x - 2\right)}}}\right| \right)} - \int{\frac{1}{x - 1} d x}$$
令 $$$u=x - 1$$$。
則 $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。
該積分可改寫為
$$\ln{\left(\left|{x - 2}\right| \right)} - {\color{red}{\int{\frac{1}{x - 1} d x}}} = \ln{\left(\left|{x - 2}\right| \right)} - {\color{red}{\int{\frac{1}{u} d u}}}$$
$$$\frac{1}{u}$$$ 的積分是 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$：
$$\ln{\left(\left|{x - 2}\right| \right)} - {\color{red}{\int{\frac{1}{u} d u}}} = \ln{\left(\left|{x - 2}\right| \right)} - {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
回顧一下 $$$u=x - 1$$$：
$$\ln{\left(\left|{x - 2}\right| \right)} - \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{x - 2}\right| \right)} - \ln{\left(\left|{{\color{red}{\left(x - 1\right)}}}\right| \right)}$$
因此，
$$\int{\frac{1}{\left(x - 2\right) \left(x - 1\right)} d x} = \ln{\left(\left|{x - 2}\right| \right)} - \ln{\left(\left|{x - 1}\right| \right)}$$
加上積分常數：
$$\int{\frac{1}{\left(x - 2\right) \left(x - 1\right)} d x} = \ln{\left(\left|{x - 2}\right| \right)} - \ln{\left(\left|{x - 1}\right| \right)}+C$$
答案
$$$\int \frac{1}{\left(x - 2\right) \left(x - 1\right)}\, dx = \left(\ln\left(\left|{x - 2}\right|\right) - \ln\left(\left|{x - 1}\right|\right)\right) + C$$$A