$$$\frac{1}{\left(x - 1\right)^{5}}$$$ 的积分

求$$$\int \frac{1}{\left(x - 1\right)^{5}}\, dx$$$。

设$$$u=x - 1$$$。

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

所以，

$${\color{red}{\int{\frac{1}{\left(x - 1\right)^{5}} d x}}} = {\color{red}{\int{\frac{1}{u^{5}} d u}}}$$

应用幂法则 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，其中 $$$n=-5$$$：

$${\color{red}{\int{\frac{1}{u^{5}} d u}}}={\color{red}{\int{u^{-5} d u}}}={\color{red}{\frac{u^{-5 + 1}}{-5 + 1}}}={\color{red}{\left(- \frac{u^{-4}}{4}\right)}}={\color{red}{\left(- \frac{1}{4 u^{4}}\right)}}$$

回忆一下 $$$u=x - 1$$$:

$$- \frac{{\color{red}{u}}^{-4}}{4} = - \frac{{\color{red}{\left(x - 1\right)}}^{-4}}{4}$$

因此，

$$\int{\frac{1}{\left(x - 1\right)^{5}} d x} = - \frac{1}{4 \left(x - 1\right)^{4}}$$

加上积分常数：

$$\int{\frac{1}{\left(x - 1\right)^{5}} d x} = - \frac{1}{4 \left(x - 1\right)^{4}}+C$$

$$$\int \frac{1}{\left(x - 1\right)^{5}}\, dx = - \frac{1}{4 \left(x - 1\right)^{4}} + C$$$A