$$$x - \frac{1}{\left(x - 1\right)^{2}}$$$ 的積分

求$$$\int \left(x - \frac{1}{\left(x - 1\right)^{2}}\right)\, dx$$$。

逐項積分：

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

套用冪次法則 $$$\int x^{n}\, dx = \frac{x^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=1$$$：

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

令 $$$u=x - 1$$$。

則 $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (步驟見»)，並可得 $$$dx = du$$$。

因此，

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

套用冪次法則 $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$，以 $$$n=-2$$$：

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

回顧一下 $$$u=x - 1$$$：

$$\frac{x^{2}}{2} + {\color{red}{u}}^{-1} = \frac{x^{2}}{2} + {\color{red}{\left(x - 1\right)}}^{-1}$$

因此，

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

化簡：

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

加上積分常數：

$$\int{\left(x - \frac{1}{\left(x - 1\right)^{2}}\right)d x} = \frac{x^{2} \left(x - 1\right) + 2}{2 \left(x - 1\right)}+C$$

$$$\int \left(x - \frac{1}{\left(x - 1\right)^{2}}\right)\, dx = \frac{x^{2} \left(x - 1\right) + 2}{2 \left(x - 1\right)} + C$$$A