1. Simplify the trigonometric expression:
Using the identities \( 1 + \cos 2x = 2\cos^2 x \) and \( \sin 2x = 2\sin x \cos x \), the numerator simplifies as follows:
\[
\frac{2 + \sin 2x}{1 + \cos 2x} = \frac{2 + 2\sin x \cos x}{2\cos^2 x} = \frac{1}{\cos^2 x} + \frac{\sin x}{\cos x} = \sec^2 x + \tan x.
\]
2. Rewrite the integral:
The integral becomes:
\[
\int \frac{2 + \sin 2x}{1 + \cos 2x} e^x \, dx = \int (\sec^2 x + \tan x) e^x \, dx.
\]
3. Separate the terms:
The integral can be split into two parts:
\[
\int (\sec^2 x + \tan x) e^x \, dx = \int \sec^2 x e^x \, dx + \int \tan x e^x \, dx.
\]
4. Solve the integrals:
The first integral yields:
\[
\int \sec^2 x e^x \, dx = e^x \tan x + C_1.
\]
The second integral yields:
\[
\int \tan x e^x \, dx = e^x \ln|\sec x| + C_2.
\]
5. Combine the results:
Summing the results of the two integrals gives:
\[
\int \frac{2 + \sin 2x}{1 + \cos 2x} e^x \, dx = e^x (\tan x + \ln|\sec x|) + C.
\]
Final Answer:
\[
\boxed{e^x (\tan x + \ln|\sec x|) + C.}
\]