How to integrate $a^{-x}$

Question

How to integrate $a^{-x}$.
This is from a text book:

$\int\frac{1}{a^x}dx, \text{a is a constant} $

I really can't think of a way of doing this, but the book says it converges using integral test, not sure what the procedures are. Thanks.

Answer 1

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

Now, $\displaystyle\frac{d(e^{mx})}{dx}=m\cdot e^{mx}\implies \int e^{mx}\ dx=\frac{e^{mx}}m+K$ wheer $m,k$ are arbitrary constants

Can you identify $m$ here?

Answer 2

$$\int \frac{dx}{a^x}=\int \exp(-x\log a)dx=-\frac{1}{a^x\log a}+C$$