Integral of $$$3^{x - 1}$$$

Find $$$\int 3^{x - 1}\, dx$$$.

Let $$$u=x - 1$$$.

Then $$$du=\left(x - 1\right)^{\prime }dx = 1 dx$$$ (steps can be seen »), and we have that $$$dx = du$$$.

So,

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

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

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

Recall that $$$u=x - 1$$$:

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

Therefore,

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

Add the constant of integration:

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

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