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We start by simplifying the integrand using trigonometric identities. Recall that $\sin 2x = 2 \sin x \cos x$ and $1 + \cos 2x = 2 \cos^2 x$. Substituting these into the integral, we get: $$ \int \frac{2 + \sin 2x}{1 + \cos 2x} e^x dx = \int \frac{2 + 2 \sin x \cos x}{2 \cos^2 x} e^x dx $$ $$ = \int \frac{1 + \sin x \cos x}{\cos^2 x} e^x dx = \int \left( \frac{1}{\cos^2 x} + \frac{\sin x \cos x}{\cos^2 x} \right) e^x dx $$ $$ = \int (\sec^2 x + \tan x) e^x dx $$
Now, we have the integral in the form $\int (\tan x + \sec^2 x) e^x dx$. We can rewrite this as $\int e^x (\tan x + \sec^2 x) dx$. Notice that $\frac{d}{dx} (\tan x) = \sec^2 x$. This suggests that we can use the formula $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$. In our case, $f(x) = \tan x$ and $f'(x) = \sec^2 x$. Therefore, $$ \int e^x (\tan x + \sec^2 x) dx = e^x \tan x + C $$
Thus, the integral evaluates to: $$ \int \frac{2 + \sin 2x}{1 + \cos 2x} e^x dx = e^x \tan x + C $$
Final Answer: $e^x \tan x + C$
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