1. The problem is to show that $$\int a^x \, dx = \frac{a^x}{\ln a} + C$$ where $a > 0$ and $a \neq 1$. 2. Recall the formula for the integral of an exponential function with base $a$: $$\int a^x \, dx = \frac{a^x}{\ln a} + C$$ where $\ln a$ is the natural logarithm of $a$. 3. To prove this, start by expressing $a^x$ in terms of the natural exponential function: $$a^x = e^{x \ln a}$$ 4. Substitute into the integral: $$\int a^x \, dx = \int e^{x \ln a} \, dx$$ 5. Use the substitution rule for integration. Let $u = x \ln a$, then $$\frac{du}{dx} = \ln a \implies dx = \frac{du}{\ln a}$$ 6. Rewrite the integral in terms of $u$: $$\int e^u \frac{du}{\ln a} = \frac{1}{\ln a} \int e^u \, du$$ 7. The integral of $e^u$ with respect to $u$ is $e^u + C$, so: $$\frac{1}{\ln a} (e^u + C) = \frac{e^u}{\ln a} + C$$ 8. Substitute back $u = x \ln a$: $$\frac{e^{x \ln a}}{\ln a} + C = \frac{a^x}{\ln a} + C$$ 9. Therefore, we have shown that $$\int a^x \, dx = \frac{a^x}{\ln a} + C$$ as required. This completes the proof.