$$$\frac{\sqrt{1 - x}}{x}$$$ 的积分

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

设$$$u=\sqrt{1 - x}$$$。

则$$$du=\left(\sqrt{1 - x}\right)^{\prime }dx = - \frac{1}{2 \sqrt{1 - x}} dx$$$ (步骤见»)，并有$$$\frac{dx}{\sqrt{1 - x}} = - 2 du$$$。

因此，

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

对 $$$c=-2$$$ 和 $$$f{\left(u \right)} = \frac{u^{2}}{1 - u^{2}}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

由于分子次数不小于分母次数，进行多项式长除法（步骤见»）:

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

逐项积分：

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

应用常数法则 $$$\int c\, du = c u$$$，使用 $$$c=1$$$：

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

进行部分分式分解（步骤可见»）:

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

逐项积分：

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

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u + 1}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

设$$$v=u + 1$$$。

则$$$dv=\left(u + 1\right)^{\prime }du = 1 du$$$ (步骤见»)，并有$$$du = dv$$$。

该积分可以改写为

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

$$$\frac{1}{v}$$$ 的积分为 $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

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

回忆一下 $$$v=u + 1$$$:

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

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \frac{1}{u - 1}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

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

设$$$v=u - 1$$$。

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

所以，

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

$$$\frac{1}{v}$$$ 的积分为 $$$\int{\frac{1}{v} d v} = \ln{\left(\left|{v}\right| \right)}$$$:

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

回忆一下 $$$v=u - 1$$$:

$$2 u - \ln{\left(\left|{u + 1}\right| \right)} + \ln{\left(\left|{{\color{red}{v}}}\right| \right)} = 2 u - \ln{\left(\left|{u + 1}\right| \right)} + \ln{\left(\left|{{\color{red}{\left(u - 1\right)}}}\right| \right)}$$

回忆一下 $$$u=\sqrt{1 - x}$$$:

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

因此，

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

加上积分常数：

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

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