Given the integral:\[I = \int e^x \frac{2 + \sin 2x}{1 + \cos 2x} \, dx.\]Step 1: Simplify the integrand. Using the identity $1 + \cos 2x = 2\cos^2 x$, we rewrite the integral as:\[I = \int e^x \frac{2 + \sin 2x}{2\cos^2 x} \, dx.\]Separating the terms and simplifying yields:\[I = \int e^x \sec^2 x \, dx + \int e^x \tan x \sec x \, dx.\]Step 2: Evaluate each integral. The first integral is $\int e^x \sec^2 x \, dx$. Using the standard result, this evaluates to $e^x \tan x + C_1$.The second integral is $\int e^x \tan x \sec x \, dx$. This evaluates to $e^x \tan x + C_2$.Step 3: Combine the results. Adding the results of the two integrals gives:\[I = e^x \tan x + C.\] Final Answer:\[\boxed{e^x \tan x + C}\]