$$$\frac{\sqrt{x^{2} + 1}}{x}$$$ 的積分

求$$$\int \frac{\sqrt{x^{2} + 1}}{x}\, dx$$$。

令 $$$x=\sinh{\left(u \right)}$$$。

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

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

因此，

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

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

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

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

積分變為

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

將分子與分母同乘以雙曲正弦，並將其餘部分用雙曲餘弦表示，使用公式 $$$\sinh^2\left(\alpha \right)=\cosh^2\left(\alpha \right)-1$$$，其中 $$$\alpha= u $$$:

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

令 $$$v=\cosh{\left(u \right)}$$$。

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

因此，

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

重寫並拆分分式:

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

逐項積分：

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

配合 $$$c=1$$$，應用常數法則 $$$\int c\, dv = c v$$$：

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

進行部分分式分解（步驟可見 »）:

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

逐項積分：

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

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 $$$f{\left(v \right)} = \frac{1}{v - 1}$$$：

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

令 $$$w=v - 1$$$。

則 $$$dw=\left(v - 1\right)^{\prime }dv = 1 dv$$$ (步驟見»)，並可得 $$$dv = dw$$$。

該積分變為

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

$$$\frac{1}{w}$$$ 的積分是 $$$\int{\frac{1}{w} d w} = \ln{\left(\left|{w}\right| \right)}$$$：

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

回顧一下 $$$w=v - 1$$$：

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

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 $$$f{\left(v \right)} = \frac{1}{v + 1}$$$：

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

令 $$$w=v + 1$$$。

則 $$$dw=\left(v + 1\right)^{\prime }dv = 1 dv$$$ (步驟見»)，並可得 $$$dv = dw$$$。

該積分變為

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

$$$\frac{1}{w}$$$ 的積分是 $$$\int{\frac{1}{w} d w} = \ln{\left(\left|{w}\right| \right)}$$$：

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

回顧一下 $$$w=v + 1$$$：

$$v + \frac{\ln{\left(\left|{v - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{w}}}\right| \right)}}{2} = v + \frac{\ln{\left(\left|{v - 1}\right| \right)}}{2} - \frac{\ln{\left(\left|{{\color{red}{\left(v + 1\right)}}}\right| \right)}}{2}$$

回顧一下 $$$v=\cosh{\left(u \right)}$$$：

$$\frac{\ln{\left(\left|{-1 + {\color{red}{v}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{1 + {\color{red}{v}}}\right| \right)}}{2} + {\color{red}{v}} = \frac{\ln{\left(\left|{-1 + {\color{red}{\cosh{\left(u \right)}}}}\right| \right)}}{2} - \frac{\ln{\left(\left|{1 + {\color{red}{\cosh{\left(u \right)}}}}\right| \right)}}{2} + {\color{red}{\cosh{\left(u \right)}}}$$

回顧一下 $$$u=\operatorname{asinh}{\left(x \right)}$$$：

$$\frac{\ln{\left(\left|{-1 + \cosh{\left({\color{red}{u}} \right)}}\right| \right)}}{2} - \frac{\ln{\left(1 + \cosh{\left({\color{red}{u}} \right)} \right)}}{2} + \cosh{\left({\color{red}{u}} \right)} = \frac{\ln{\left(\left|{-1 + \cosh{\left({\color{red}{\operatorname{asinh}{\left(x \right)}}} \right)}}\right| \right)}}{2} - \frac{\ln{\left(1 + \cosh{\left({\color{red}{\operatorname{asinh}{\left(x \right)}}} \right)} \right)}}{2} + \cosh{\left({\color{red}{\operatorname{asinh}{\left(x \right)}}} \right)}$$

因此，

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

加上積分常數：

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

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