$$$\frac{\sqrt{9 - x^{2}}}{x^{2}}$$$ 的积分

求$$$\int \frac{\sqrt{9 - x^{2}}}{x^{2}}\, dx$$$。

设$$$x=3 \sin{\left(u \right)}$$$。

则$$$dx=\left(3 \sin{\left(u \right)}\right)^{\prime }du = 3 \cos{\left(u \right)} du$$$（步骤见»）。

此外，可得$$$u=\operatorname{asin}{\left(\frac{x}{3} \right)}$$$。

因此，

$$$\frac{\sqrt{9 - x^{2}}}{x^{2}} = \frac{\sqrt{9 - 9 \sin^{2}{\left( u \right)}}}{9 \sin^{2}{\left( u \right)}}$$$

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

$$$\frac{\sqrt{9 - 9 \sin^{2}{\left( u \right)}}}{9 \sin^{2}{\left( u \right)}}=\frac{\sqrt{1 - \sin^{2}{\left( u \right)}}}{3 \sin^{2}{\left( u \right)}}=\frac{\sqrt{\cos^{2}{\left( u \right)}}}{3 \sin^{2}{\left( u \right)}}$$$

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

$$$\frac{\sqrt{\cos^{2}{\left( u \right)}}}{3 \sin^{2}{\left( u \right)}} = \frac{\cos{\left( u \right)}}{3 \sin^{2}{\left( u \right)}}$$$

因此，

$${\color{red}{\int{\frac{\sqrt{9 - x^{2}}}{x^{2}} d x}}} = {\color{red}{\int{\frac{\cos^{2}{\left(u \right)}}{\sin^{2}{\left(u \right)}} d u}}}$$

用余切表示:

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

设$$$v=\cot{\left(u \right)}$$$。

则$$$dv=\left(\cot{\left(u \right)}\right)^{\prime }du = - \csc^{2}{\left(u \right)} du$$$ (步骤见»)，并有$$$\csc^{2}{\left(u \right)} du = - dv$$$。

该积分可以改写为

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

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

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

改写并拆分该分式:

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

逐项积分：

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

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

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

$$$\frac{1}{v^{2} + 1}$$$ 的积分为 $$$\int{\frac{1}{v^{2} + 1} d v} = \operatorname{atan}{\left(v \right)}$$$:

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

回忆一下 $$$v=\cot{\left(u \right)}$$$:

$$\operatorname{atan}{\left({\color{red}{v}} \right)} - {\color{red}{v}} = \operatorname{atan}{\left({\color{red}{\cot{\left(u \right)}}} \right)} - {\color{red}{\cot{\left(u \right)}}}$$

回忆一下 $$$u=\operatorname{asin}{\left(\frac{x}{3} \right)}$$$:

$$- \cot{\left({\color{red}{u}} \right)} + \operatorname{atan}{\left(\cot{\left({\color{red}{u}} \right)} \right)} = - \cot{\left({\color{red}{\operatorname{asin}{\left(\frac{x}{3} \right)}}} \right)} + \operatorname{atan}{\left(\cot{\left({\color{red}{\operatorname{asin}{\left(\frac{x}{3} \right)}}} \right)} \right)}$$

因此，

$$\int{\frac{\sqrt{9 - x^{2}}}{x^{2}} d x} = \operatorname{atan}{\left(\frac{3 \sqrt{1 - \frac{x^{2}}{9}}}{x} \right)} - \frac{3 \sqrt{1 - \frac{x^{2}}{9}}}{x}$$

化简：

$$\int{\frac{\sqrt{9 - x^{2}}}{x^{2}} d x} = \operatorname{atan}{\left(\frac{\sqrt{9 - x^{2}}}{x} \right)} - \frac{\sqrt{9 - x^{2}}}{x}$$

加上积分常数：

$$\int{\frac{\sqrt{9 - x^{2}}}{x^{2}} d x} = \operatorname{atan}{\left(\frac{\sqrt{9 - x^{2}}}{x} \right)} - \frac{\sqrt{9 - x^{2}}}{x}+C$$

$$$\int \frac{\sqrt{9 - x^{2}}}{x^{2}}\, dx = \left(\operatorname{atan}{\left(\frac{\sqrt{9 - x^{2}}}{x} \right)} - \frac{\sqrt{9 - x^{2}}}{x}\right) + C$$$A