$$$\frac{\sqrt{x - 1}}{x}$$$ 的積分

求$$$\int \frac{\sqrt{x - 1}}{x}\, dx$$$。

令 $$$u=\sqrt{x - 1}$$$。

則 $$$du=\left(\sqrt{x - 1}\right)^{\prime }dx = \frac{1}{2 \sqrt{x - 1}} dx$$$ (步驟見»)，並可得 $$$\frac{dx}{\sqrt{x - 1}} = 2 du$$$。

所以，

$${\color{red}{\int{\frac{\sqrt{x - 1}}{x} d x}}} = {\color{red}{\int{\frac{2 u^{2}}{u^{2} + 1} d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=2$$$ 與 $$$f{\left(u \right)} = \frac{u^{2}}{u^{2} + 1}$$$：

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

重寫並拆分分式:

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

逐項積分：

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

配合 $$$c=1$$$，應用常數法則 $$$\int c\, du = c u$$$：

$$- 2 \int{\frac{1}{u^{2} + 1} d u} + 2 {\color{red}{\int{1 d u}}} = - 2 \int{\frac{1}{u^{2} + 1} d u} + 2 {\color{red}{u}}$$

$$$\frac{1}{u^{2} + 1}$$$ 的積分是 $$$\int{\frac{1}{u^{2} + 1} d u} = \operatorname{atan}{\left(u \right)}$$$：

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

回顧一下 $$$u=\sqrt{x - 1}$$$：

$$- 2 \operatorname{atan}{\left({\color{red}{u}} \right)} + 2 {\color{red}{u}} = - 2 \operatorname{atan}{\left({\color{red}{\sqrt{x - 1}}} \right)} + 2 {\color{red}{\sqrt{x - 1}}}$$

因此，

$$\int{\frac{\sqrt{x - 1}}{x} d x} = 2 \sqrt{x - 1} - 2 \operatorname{atan}{\left(\sqrt{x - 1} \right)}$$

加上積分常數：

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

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