What is the integral of $a^{x-1}$?

Question

What is the integral of $$\int a^{x-1}dx?$$

is it $$\frac{a^{x-1}}{\log(a)} + c?$$

How can we derive the proper integral? Also can you please tell me the definite integral with limits, say b to c?

Answer 1

it is $$\frac{1}{a}\int a^xdx=\frac{1}{a}\frac{a^x}{\ln(a)}+C$$

Answer 2

Let $a>0$, $a,x$ real:

$a^{x-1} =\exp(\ln(a^{x-1}))= $

$\exp((x-1)\ln(a))= \exp(c(x-1))$,

where $c :=\ln(a).$

Use substitution:

$y=c (x-1)$ to integrate $\exp(y).$