$$$\sqrt{a^{x} - 1}$$$ 关于$$$x$$$的积分

求$$$\int \sqrt{a^{x} - 1}\, dx$$$。

设$$$u=\sqrt{a^{x} - 1}$$$。

则$$$du=\left(\sqrt{a^{x} - 1}\right)^{\prime }dx = \frac{a^{x} \ln{\left(a \right)}}{2 \sqrt{a^{x} - 1}} dx$$$ (步骤见»)，并有$$$\frac{a^{x} dx}{\sqrt{a^{x} - 1}} = \frac{2 du}{\ln{\left(a \right)}}$$$。

所以，

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

对 $$$c=\frac{2}{\ln{\left(a \right)}}$$$ 和 $$$f{\left(u \right)} = \frac{u^{2}}{u^{2} + 1}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$${\color{red}{\int{\frac{2 u^{2}}{\left(u^{2} + 1\right) \ln{\left(a \right)}} d u}}} = {\color{red}{\left(\frac{2 \int{\frac{u^{2}}{u^{2} + 1} d u}}{\ln{\left(a \right)}}\right)}}$$

改写并拆分该分式:

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

逐项积分：

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

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

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

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

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

回忆一下 $$$u=\sqrt{a^{x} - 1}$$$:

$$\frac{2 \left(- \operatorname{atan}{\left({\color{red}{u}} \right)} + {\color{red}{u}}\right)}{\ln{\left(a \right)}} = \frac{2 \left(- \operatorname{atan}{\left({\color{red}{\sqrt{a^{x} - 1}}} \right)} + {\color{red}{\sqrt{a^{x} - 1}}}\right)}{\ln{\left(a \right)}}$$

因此，

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

加上积分常数：

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

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