$$$\frac{\sqrt{x^{2} - 100}}{x}$$$ 的积分

求$$$\int \frac{\sqrt{x^{2} - 100}}{x}\, dx$$$。

设$$$x=10 \cosh{\left(u \right)}$$$。

则$$$dx=\left(10 \cosh{\left(u \right)}\right)^{\prime }du = 10 \sinh{\left(u \right)} du$$$（步骤见»）。

此外，可得$$$u=\operatorname{acosh}{\left(\frac{x}{10} \right)}$$$。

被积函数变为

$$$\frac{\sqrt{x^{2} - 100}}{x} = \frac{\sqrt{100 \cosh^{2}{\left( u \right)} - 100}}{10 \cosh{\left( u \right)}}$$$

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

$$$\frac{\sqrt{100 \cosh^{2}{\left( u \right)} - 100}}{10 \cosh{\left( u \right)}}=\frac{\sqrt{\cosh^{2}{\left( u \right)} - 1}}{\cosh{\left( u \right)}}=\frac{\sqrt{\sinh^{2}{\left( u \right)}}}{\cosh{\left( u \right)}}$$$

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

$$$\frac{\sqrt{\sinh^{2}{\left( u \right)}}}{\cosh{\left( u \right)}} = \frac{\sinh{\left( u \right)}}{\cosh{\left( u \right)}}$$$

因此，

$${\color{red}{\int{\frac{\sqrt{x^{2} - 100}}{x} d x}}} = {\color{red}{\int{\frac{10 \sinh^{2}{\left(u \right)}}{\cosh{\left(u \right)}} d u}}}$$

对 $$$c=10$$$ 和 $$$f{\left(u \right)} = \frac{\sinh^{2}{\left(u \right)}}{\cosh{\left(u \right)}}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\frac{10 \sinh^{2}{\left(u \right)}}{\cosh{\left(u \right)}} d u}}} = {\color{red}{\left(10 \int{\frac{\sinh^{2}{\left(u \right)}}{\cosh{\left(u \right)}} d u}\right)}}$$

将分子和分母同乘以一个双曲余弦，并使用公式 $$$\cosh^2\left(\alpha \right)=\sinh^2\left(\alpha \right)+1$$$（其中 $$$\alpha= u $$$），把其余全部用双曲正弦表示。:

$$10 {\color{red}{\int{\frac{\sinh^{2}{\left(u \right)}}{\cosh{\left(u \right)}} d u}}} = 10 {\color{red}{\int{\frac{\sinh^{2}{\left(u \right)} \cosh{\left(u \right)}}{\sinh^{2}{\left(u \right)} + 1} d u}}}$$

设$$$v=\sinh{\left(u \right)}$$$。

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

积分变为

$$10 {\color{red}{\int{\frac{\sinh^{2}{\left(u \right)} \cosh{\left(u \right)}}{\sinh^{2}{\left(u \right)} + 1} d u}}} = 10 {\color{red}{\int{\frac{v^{2}}{v^{2} + 1} d v}}}$$

改写并拆分该分式:

$$10 {\color{red}{\int{\frac{v^{2}}{v^{2} + 1} d v}}} = 10 {\color{red}{\int{\left(1 - \frac{1}{v^{2} + 1}\right)d v}}}$$

逐项积分：

$$10 {\color{red}{\int{\left(1 - \frac{1}{v^{2} + 1}\right)d v}}} = 10 {\color{red}{\left(\int{1 d v} - \int{\frac{1}{v^{2} + 1} d v}\right)}}$$

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

$$- 10 \int{\frac{1}{v^{2} + 1} d v} + 10 {\color{red}{\int{1 d v}}} = - 10 \int{\frac{1}{v^{2} + 1} d v} + 10 {\color{red}{v}}$$

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

$$10 v - 10 {\color{red}{\int{\frac{1}{v^{2} + 1} d v}}} = 10 v - 10 {\color{red}{\operatorname{atan}{\left(v \right)}}}$$

回忆一下 $$$v=\sinh{\left(u \right)}$$$:

$$- 10 \operatorname{atan}{\left({\color{red}{v}} \right)} + 10 {\color{red}{v}} = - 10 \operatorname{atan}{\left({\color{red}{\sinh{\left(u \right)}}} \right)} + 10 {\color{red}{\sinh{\left(u \right)}}}$$

回忆一下 $$$u=\operatorname{acosh}{\left(\frac{x}{10} \right)}$$$:

$$10 \sinh{\left({\color{red}{u}} \right)} - 10 \operatorname{atan}{\left(\sinh{\left({\color{red}{u}} \right)} \right)} = 10 \sinh{\left({\color{red}{\operatorname{acosh}{\left(\frac{x}{10} \right)}}} \right)} - 10 \operatorname{atan}{\left(\sinh{\left({\color{red}{\operatorname{acosh}{\left(\frac{x}{10} \right)}}} \right)} \right)}$$

因此，

$$\int{\frac{\sqrt{x^{2} - 100}}{x} d x} = 10 \sqrt{\frac{x}{10} - 1} \sqrt{\frac{x}{10} + 1} - 10 \operatorname{atan}{\left(\sqrt{\frac{x}{10} - 1} \sqrt{\frac{x}{10} + 1} \right)}$$

化简：

$$\int{\frac{\sqrt{x^{2} - 100}}{x} d x} = \sqrt{x - 10} \sqrt{x + 10} - 10 \operatorname{atan}{\left(\frac{\sqrt{x - 10} \sqrt{x + 10}}{10} \right)}$$

加上积分常数：

$$\int{\frac{\sqrt{x^{2} - 100}}{x} d x} = \sqrt{x - 10} \sqrt{x + 10} - 10 \operatorname{atan}{\left(\frac{\sqrt{x - 10} \sqrt{x + 10}}{10} \right)}+C$$

$$$\int \frac{\sqrt{x^{2} - 100}}{x}\, dx = \left(\sqrt{x - 10} \sqrt{x + 10} - 10 \operatorname{atan}{\left(\frac{\sqrt{x - 10} \sqrt{x + 10}}{10} \right)}\right) + C$$$A