# Integral of $a^x$

 

Evaluate the integral $\int a^x \; dx$ where $a$ is a constant such that $a \gt 0$ and $a \ne 1$

We first change the base of the exponential $a^x$

Let $y = a^x$ and take the $\ln$ of both sides
$\qquad \ln y = \ln (a^x)$

The property of $\; \ln \;$ given by $\; \ln a^x = x \ln a \;$ in used to write the above as
$\qquad \ln y = x \ln a$

Use the relationship between the logarithm and the exponential to write the above as
$y = e^{x \ln a}$
and since $y = a^x$, we obtain
$a^x = e^{x \ln a}$
Using the above, the given integral may be written as
$\qquad \displaystyle \int a^x \; dx = \int e^{x \ln a} dx \qquad (1)$

We now evaluate the integral on the right side
Let $u(x) = x \ln a$ which gives $\dfrac{du}{dx} = \ln a$

By definition,
1 , 2 , 3 the differential $du$ is given by $du = \dfrac{du}{dx} dx = \ln a \; dx$, hence $du = \ln a \; dx$

Use the substitution $u(x) = x \; \ln a$ in the right integral in $(1)$ and write
$\qquad \displaystyle \int e^{x \ln a} dx = \int e^{u} \; dx$

Divide and multiply the integrand on the right by $\ln a$
$\qquad \displaystyle \int e^{x \ln a} dx = \int \dfrac{1}{\ln a}e^{u} \ln a \; dx$

Use the fact that $du = \ln a \; dx$ in the above and rewrite the integral as
$\qquad \displaystyle \int e^{x \ln a} dx = \int \dfrac{1}{\ln a}e^{u} \; du$

Evalute the above integral using the
integral formula $\displaystyle \int e^x dx = e^x + c$ to obtain
$\qquad \displaystyle \int e^{x \ln a} dx = \dfrac{1}{\ln a} e^u + c$ , where $c$ is the constant on integration.

Substitute back $u$ and write
$\qquad \displaystyle \int e^{x \ln a} dx = \dfrac{1}{\ln a}e^{x \ln a} + c$

and finally using $\displaystyle \int a^x \; dx = \int e^{x \ln a} dx$ in $(1)$ above , we write
$\int a^x \; dx = \dfrac{1}{\ln a}e^{x \ln a} + c$ or using the fact that $a^x = e^{x \ln a}$, $\int a^x \; dx = \dfrac{1}{\ln a} a^x + c$