1. **State the problem:** We need to evaluate the integral $$\int \frac{e^{2x}}{e^x + 3} \, dx.$$\n\n2. **Rewrite the integral:** Notice that $e^{2x} = (e^x)^2$. So the integral becomes $$\int \frac{(e^x)^2}{e^x + 3} \, dx.$$\n\n3. **Use substitution:** Let $u = e^x + 3$. Then, $$\frac{du}{dx} = e^x \implies du = e^x \, dx.$$\n\n4. **Express $e^{2x} dx$ in terms of $u$ and $du$:** Since $e^{2x} dx = e^x \cdot e^x dx = e^x (e^x dx)$, and from substitution $e^x dx = du$, we have $$e^{2x} dx = e^x \cdot du = (u - 3) du,$$ because $u = e^x + 3 \implies e^x = u - 3.$\n\n5. **Rewrite the integral in terms of $u$:** The integral becomes $$\int \frac{e^{2x}}{e^x + 3} \, dx = \int \frac{(u - 3) du}{u} = \int \left(1 - \frac{3}{u}\right) du.$$\n\n6. **Integrate term-by-term:** $$\int 1 \, du - 3 \int \frac{1}{u} \, du = u - 3 \ln|u| + C,$$ where $C$ is the constant of integration.\n\n7. **Back-substitute $u = e^x + 3$:** $$\boxed{e^x + 3 - 3 \ln|e^x + 3| + C}.$$\n\nThis is the final answer.