$$$\frac{\sqrt{x}}{\sqrt{1 - x}}$$$ 的积分

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

设$$$u=\sqrt{x}$$$。

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

因此，

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

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

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

设$$$u=\sin{\left(v \right)}$$$。

则$$$du=\left(\sin{\left(v \right)}\right)^{\prime }dv = \cos{\left(v \right)} dv$$$（步骤见»）。

此外，可得$$$v=\operatorname{asin}{\left(u \right)}$$$。

因此，

$$$\frac{ u ^{2}}{\sqrt{1 - u ^{2}}} = \frac{\sin^{2}{\left( v \right)}}{\sqrt{1 - \sin^{2}{\left( v \right)}}}$$$

利用恒等式 $$$1 - \sin^{2}{\left( v \right)} = \cos^{2}{\left( v \right)}$$$：

$$$\frac{\sin^{2}{\left( v \right)}}{\sqrt{1 - \sin^{2}{\left( v \right)}}}=\frac{\sin^{2}{\left( v \right)}}{\sqrt{\cos^{2}{\left( v \right)}}}$$$

假设$$$\cos{\left( v \right)} \ge 0$$$，我们得到如下结果：

$$$\frac{\sin^{2}{\left( v \right)}}{\sqrt{\cos^{2}{\left( v \right)}}} = \frac{\sin^{2}{\left( v \right)}}{\cos{\left( v \right)}}$$$

所以，

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

应用降幂公式 $$$\sin^{2}{\left(\alpha \right)} = \frac{1}{2} - \frac{\cos{\left(2 \alpha \right)}}{2}$$$，并令 $$$\alpha= v $$$:

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

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

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

逐项积分：

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

应用常数法则 $$$\int c\, dv = c v$$$，使用 $$$c=1$$$：

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

设$$$w=2 v$$$。

则$$$dw=\left(2 v\right)^{\prime }dv = 2 dv$$$ (步骤见»)，并有$$$dv = \frac{dw}{2}$$$。

该积分可以改写为

$$v - {\color{red}{\int{\cos{\left(2 v \right)} d v}}} = v - {\color{red}{\int{\frac{\cos{\left(w \right)}}{2} d w}}}$$

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

$$v - {\color{red}{\int{\frac{\cos{\left(w \right)}}{2} d w}}} = v - {\color{red}{\left(\frac{\int{\cos{\left(w \right)} d w}}{2}\right)}}$$

余弦函数的积分为 $$$\int{\cos{\left(w \right)} d w} = \sin{\left(w \right)}$$$：

$$v - \frac{{\color{red}{\int{\cos{\left(w \right)} d w}}}}{2} = v - \frac{{\color{red}{\sin{\left(w \right)}}}}{2}$$

回忆一下 $$$w=2 v$$$:

$$v - \frac{\sin{\left({\color{red}{w}} \right)}}{2} = v - \frac{\sin{\left({\color{red}{\left(2 v\right)}} \right)}}{2}$$

回忆一下 $$$v=\operatorname{asin}{\left(u \right)}$$$:

$$- \frac{\sin{\left(2 {\color{red}{v}} \right)}}{2} + {\color{red}{v}} = - \frac{\sin{\left(2 {\color{red}{\operatorname{asin}{\left(u \right)}}} \right)}}{2} + {\color{red}{\operatorname{asin}{\left(u \right)}}}$$

回忆一下 $$$u=\sqrt{x}$$$:

$$- \frac{\sin{\left(2 \operatorname{asin}{\left({\color{red}{u}} \right)} \right)}}{2} + \operatorname{asin}{\left({\color{red}{u}} \right)} = - \frac{\sin{\left(2 \operatorname{asin}{\left({\color{red}{\sqrt{x}}} \right)} \right)}}{2} + \operatorname{asin}{\left({\color{red}{\sqrt{x}}} \right)}$$

因此，

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

使用公式 $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$，化简该表达式：

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

加上积分常数：

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

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