$$$\sqrt{\frac{1 - x}{x + 1}}$$$ 的积分

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

输入已重写为：$$$\int{\sqrt{\frac{1 - x}{x + 1}} d x}=\int{\frac{\sqrt{1 - x}}{\sqrt{x + 1}} d x}$$$。

将分子和分母都乘以 $$$\sqrt{x + 1}$$$ 并化简:

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

设$$$x=\sin{\left(u \right)}$$$。

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

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

因此，

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

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

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

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

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

积分变为

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

将余弦用正弦表示，进一步改写分子，利用平方差公式，并化简。:

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

逐项积分：

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

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

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

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

$$u - {\color{red}{\int{\sin{\left(u \right)} d u}}} = u - {\color{red}{\left(- \cos{\left(u \right)}\right)}}$$

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

$$\cos{\left({\color{red}{u}} \right)} + {\color{red}{u}} = \cos{\left({\color{red}{\operatorname{asin}{\left(x \right)}}} \right)} + {\color{red}{\operatorname{asin}{\left(x \right)}}}$$

因此，

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

加上积分常数：

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

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