1. **State the problem:** Show that the integral of $a^x$ with respect to $x$ is $$\int a^x \, dx = \frac{e^{x \ln a}}{\ln a} + C$$ where $a > 0$ and $a \neq 1$. 2. **Recall the formula and rules:** The function $a^x$ can be rewritten using the natural exponential function as $$a^x = e^{x \ln a}$$ where $\ln a$ is the natural logarithm of $a$. 3. **Rewrite the integral:** Using this, the integral becomes $$\int a^x \, dx = \int e^{x \ln a} \, dx$$ 4. **Integrate using substitution:** Let $u = x \ln a$, then $$\frac{du}{dx} = \ln a \implies dx = \frac{du}{\ln a}$$ 5. **Substitute and integrate:** $$\int e^{x \ln a} \, dx = \int e^u \frac{du}{\ln a} = \frac{1}{\ln a} \int e^u \, du = \frac{1}{\ln a} e^u + C$$ 6. **Back-substitute $u$:** $$\frac{1}{\ln a} e^{x \ln a} + C = \frac{a^x}{\ln a} + C$$ 7. **Conclusion:** Therefore, $$\int a^x \, dx = \frac{a^x}{\ln a} + C$$ which matches the given expression. This shows the integral of $a^x$ with respect to $x$ is $$\frac{a^x}{\ln a} + C$$ as required.