$$$a^{x} - 1$$$ 對 $$$x$$$ 的積分

求$$$\int \left(a^{x} - 1\right)\, dx$$$。

逐項積分：

$${\color{red}{\int{\left(a^{x} - 1\right)d x}}} = {\color{red}{\left(- \int{1 d x} + \int{a^{x} d x}\right)}}$$

配合 $$$c=1$$$，應用常數法則 $$$\int c\, dx = c x$$$：

$$\int{a^{x} d x} - {\color{red}{\int{1 d x}}} = \int{a^{x} d x} - {\color{red}{x}}$$

Apply the exponential rule $$$\int{a^{x} d x} = \frac{a^{x}}{\ln{\left(a \right)}}$$$ with $$$a=a$$$:

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

因此，

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

加上積分常數：

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

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