Integral of $$$b^{x - 1}$$$ with respect to $$$x$$$

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

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

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

Thus,

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

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

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

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

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

Therefore,

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

Add the constant of integration:

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

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