$$$3^{x - 1}$$$ 的积分

求$$$\int 3^{x - 1}\, dx$$$。

设$$$u=x - 1$$$。

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

因此，

$${\color{red}{\int{3^{x - 1} d x}}} = {\color{red}{\int{3^{u} d u}}}$$

Apply the exponential rule $$$\int{a^{u} d u} = \frac{a^{u}}{\ln{\left(a \right)}}$$$ with $$$a=3$$$:

$${\color{red}{\int{3^{u} d u}}} = {\color{red}{\frac{3^{u}}{\ln{\left(3 \right)}}}}$$

回忆一下 $$$u=x - 1$$$:

$$\frac{3^{{\color{red}{u}}}}{\ln{\left(3 \right)}} = \frac{3^{{\color{red}{\left(x - 1\right)}}}}{\ln{\left(3 \right)}}$$

因此，

$$\int{3^{x - 1} d x} = \frac{3^{x - 1}}{\ln{\left(3 \right)}}$$

加上积分常数：

$$\int{3^{x - 1} d x} = \frac{3^{x - 1}}{\ln{\left(3 \right)}}+C$$

$$$\int 3^{x - 1}\, dx = \frac{3^{x - 1}}{\ln\left(3\right)} + C$$$A