$$$\sqrt{x^{2} - 6}$$$ 的積分

求$$$\int \sqrt{x^{2} - 6}\, dx$$$。

令 $$$x=\sqrt{6} \cosh{\left(u \right)}$$$。

則 $$$dx=\left(\sqrt{6} \cosh{\left(u \right)}\right)^{\prime }du = \sqrt{6} \sinh{\left(u \right)} du$$$（步驟見»）。

此外，由此可得 $$$u=\operatorname{acosh}{\left(\frac{\sqrt{6} x}{6} \right)}$$$。

所以，

$$$\sqrt{x^{2} - 6} = \sqrt{6 \cosh^{2}{\left( u \right)} - 6}$$$

使用恆等式 $$$\cosh^{2}{\left( u \right)} - 1 = \sinh^{2}{\left( u \right)}$$$：

$$$\sqrt{6 \cosh^{2}{\left( u \right)} - 6}=\sqrt{6} \sqrt{\cosh^{2}{\left( u \right)} - 1}=\sqrt{6} \sqrt{\sinh^{2}{\left( u \right)}}$$$

假設 $$$\sinh{\left( u \right)} \ge 0$$$，可得如下：

$$$\sqrt{6} \sqrt{\sinh^{2}{\left( u \right)}} = \sqrt{6} \sinh{\left( u \right)}$$$

積分可以改寫為

$${\color{red}{\int{\sqrt{x^{2} - 6} d x}}} = {\color{red}{\int{6 \sinh^{2}{\left(u \right)} d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=6$$$ 與 $$$f{\left(u \right)} = \sinh^{2}{\left(u \right)}$$$：

$${\color{red}{\int{6 \sinh^{2}{\left(u \right)} d u}}} = {\color{red}{\left(6 \int{\sinh^{2}{\left(u \right)} d u}\right)}}$$

套用降冪公式 $$$\sinh^{2}{\left(\alpha \right)} = \frac{\cosh{\left(2 \alpha \right)}}{2} - \frac{1}{2}$$$，令 $$$\alpha= u $$$:

$$6 {\color{red}{\int{\sinh^{2}{\left(u \right)} d u}}} = 6 {\color{red}{\int{\left(\frac{\cosh{\left(2 u \right)}}{2} - \frac{1}{2}\right)d u}}}$$

套用常數倍法則 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(u \right)} = \cosh{\left(2 u \right)} - 1$$$：

$$6 {\color{red}{\int{\left(\frac{\cosh{\left(2 u \right)}}{2} - \frac{1}{2}\right)d u}}} = 6 {\color{red}{\left(\frac{\int{\left(\cosh{\left(2 u \right)} - 1\right)d u}}{2}\right)}}$$

逐項積分：

$$3 {\color{red}{\int{\left(\cosh{\left(2 u \right)} - 1\right)d u}}} = 3 {\color{red}{\left(- \int{1 d u} + \int{\cosh{\left(2 u \right)} d u}\right)}}$$

配合 $$$c=1$$$，應用常數法則 $$$\int c\, du = c u$$$：

$$3 \int{\cosh{\left(2 u \right)} d u} - 3 {\color{red}{\int{1 d u}}} = 3 \int{\cosh{\left(2 u \right)} d u} - 3 {\color{red}{u}}$$

令 $$$v=2 u$$$。

則 $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (步驟見»)，並可得 $$$du = \frac{dv}{2}$$$。

因此，

$$- 3 u + 3 {\color{red}{\int{\cosh{\left(2 u \right)} d u}}} = - 3 u + 3 {\color{red}{\int{\frac{\cosh{\left(v \right)}}{2} d v}}}$$

套用常數倍法則 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$，使用 $$$c=\frac{1}{2}$$$ 與 $$$f{\left(v \right)} = \cosh{\left(v \right)}$$$：

$$- 3 u + 3 {\color{red}{\int{\frac{\cosh{\left(v \right)}}{2} d v}}} = - 3 u + 3 {\color{red}{\left(\frac{\int{\cosh{\left(v \right)} d v}}{2}\right)}}$$

雙曲餘弦的積分為 $$$\int{\cosh{\left(v \right)} d v} = \sinh{\left(v \right)}$$$：

$$- 3 u + \frac{3 {\color{red}{\int{\cosh{\left(v \right)} d v}}}}{2} = - 3 u + \frac{3 {\color{red}{\sinh{\left(v \right)}}}}{2}$$

回顧一下 $$$v=2 u$$$：

$$- 3 u + \frac{3 \sinh{\left({\color{red}{v}} \right)}}{2} = - 3 u + \frac{3 \sinh{\left({\color{red}{\left(2 u\right)}} \right)}}{2}$$

回顧一下 $$$u=\operatorname{acosh}{\left(\frac{\sqrt{6} x}{6} \right)}$$$：

$$\frac{3 \sinh{\left(2 {\color{red}{u}} \right)}}{2} - 3 {\color{red}{u}} = \frac{3 \sinh{\left(2 {\color{red}{\operatorname{acosh}{\left(\frac{\sqrt{6} x}{6} \right)}}} \right)}}{2} - 3 {\color{red}{\operatorname{acosh}{\left(\frac{\sqrt{6} x}{6} \right)}}}$$

因此，

$$\int{\sqrt{x^{2} - 6} d x} = \frac{3 \sinh{\left(2 \operatorname{acosh}{\left(\frac{\sqrt{6} x}{6} \right)} \right)}}{2} - 3 \operatorname{acosh}{\left(\frac{\sqrt{6} x}{6} \right)}$$

使用公式 $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$，化簡該表達式：

$$\int{\sqrt{x^{2} - 6} d x} = \frac{\sqrt{6} x \sqrt{\frac{\sqrt{6} x}{6} - 1} \sqrt{\frac{\sqrt{6} x}{6} + 1}}{2} - 3 \operatorname{acosh}{\left(\frac{\sqrt{6} x}{6} \right)}$$

進一步化簡：

$$\int{\sqrt{x^{2} - 6} d x} = \frac{\sqrt{6} x \sqrt{\sqrt{6} x - 6} \sqrt{\sqrt{6} x + 6}}{12} - 3 \operatorname{acosh}{\left(\frac{\sqrt{6} x}{6} \right)}$$

加上積分常數：

$$\int{\sqrt{x^{2} - 6} d x} = \frac{\sqrt{6} x \sqrt{\sqrt{6} x - 6} \sqrt{\sqrt{6} x + 6}}{12} - 3 \operatorname{acosh}{\left(\frac{\sqrt{6} x}{6} \right)}+C$$

$$$\int \sqrt{x^{2} - 6}\, dx = \left(\frac{\sqrt{6} x \sqrt{\sqrt{6} x - 6} \sqrt{\sqrt{6} x + 6}}{12} - 3 \operatorname{acosh}{\left(\frac{\sqrt{6} x}{6} \right)}\right) + C$$$A