$$$\frac{1}{x^{2} - 5 x + 6}$$$ 的积分
您的输入
求$$$\int \frac{1}{x^{2} - 5 x + 6}\, dx$$$。
解答
进行部分分式分解（步骤可见»）:
$${\color{red}{\int{\frac{1}{x^{2} - 5 x + 6} d x}}} = {\color{red}{\int{\left(- \frac{1}{x - 2} + \frac{1}{x - 3}\right)d x}}}$$
逐项积分：
$${\color{red}{\int{\left(- \frac{1}{x - 2} + \frac{1}{x - 3}\right)d x}}} = {\color{red}{\left(\int{\frac{1}{x - 3} d x} - \int{\frac{1}{x - 2} d x}\right)}}$$
设$$$u=x - 3$$$。
则$$$du=\left(x - 3\right)^{\prime }dx = 1 dx$$$ (步骤见»)，并有$$$dx = du$$$。
该积分可以改写为
$$- \int{\frac{1}{x - 2} d x} + {\color{red}{\int{\frac{1}{x - 3} d x}}} = - \int{\frac{1}{x - 2} d x} + {\color{red}{\int{\frac{1}{u} d u}}}$$
$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$- \int{\frac{1}{x - 2} d x} + {\color{red}{\int{\frac{1}{u} d u}}} = - \int{\frac{1}{x - 2} d x} + {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
回忆一下 $$$u=x - 3$$$:
$$\ln{\left(\left|{{\color{red}{u}}}\right| \right)} - \int{\frac{1}{x - 2} d x} = \ln{\left(\left|{{\color{red}{\left(x - 3\right)}}}\right| \right)} - \int{\frac{1}{x - 2} d x}$$
设$$$u=x - 2$$$。
则$$$du=\left(x - 2\right)^{\prime }dx = 1 dx$$$ (步骤见»)，并有$$$dx = du$$$。
所以，
$$\ln{\left(\left|{x - 3}\right| \right)} - {\color{red}{\int{\frac{1}{x - 2} d x}}} = \ln{\left(\left|{x - 3}\right| \right)} - {\color{red}{\int{\frac{1}{u} d u}}}$$
$$$\frac{1}{u}$$$ 的积分为 $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:
$$\ln{\left(\left|{x - 3}\right| \right)} - {\color{red}{\int{\frac{1}{u} d u}}} = \ln{\left(\left|{x - 3}\right| \right)} - {\color{red}{\ln{\left(\left|{u}\right| \right)}}}$$
回忆一下 $$$u=x - 2$$$:
$$\ln{\left(\left|{x - 3}\right| \right)} - \ln{\left(\left|{{\color{red}{u}}}\right| \right)} = \ln{\left(\left|{x - 3}\right| \right)} - \ln{\left(\left|{{\color{red}{\left(x - 2\right)}}}\right| \right)}$$
因此，
$$\int{\frac{1}{x^{2} - 5 x + 6} d x} = \ln{\left(\left|{x - 3}\right| \right)} - \ln{\left(\left|{x - 2}\right| \right)}$$
加上积分常数：
$$\int{\frac{1}{x^{2} - 5 x + 6} d x} = \ln{\left(\left|{x - 3}\right| \right)} - \ln{\left(\left|{x - 2}\right| \right)}+C$$
答案
$$$\int \frac{1}{x^{2} - 5 x + 6}\, dx = \left(\ln\left(\left|{x - 3}\right|\right) - \ln\left(\left|{x - 2}\right|\right)\right) + C$$$A