Integrate the function: $\frac{2+\sin 2x}{1+\cos 2x} e^x$

Let $I = \int \left( \frac{2+\sin 2x}{1+\cos 2x} \right) e^x dx$
Using trigonometric identities $\sin 2x = 2 \sin x \cos x$ and $1 + \cos 2x = 2 \cos^2 x$:
$I = \int \left( \frac{2 + 2 \sin x \cos x}{2 \cos^2 x} \right) e^x dx$
$I = \int \left( \frac{2(1 + \sin x \cos x)}{2 \cos^2 x} \right) e^x dx$
$I = \int \left( \frac{1}{\cos^2 x} + \frac{\sin x \cos x}{\cos^2 x} \right) e^x dx$
$I = \int (\sec^2 x + \tan x) e^x dx$
We know that $\int (f(x) + f'(x)) e^x dx = e^x f(x) + C$.
Here,let $f(x) = \tan x$,then $f'(x) = \sec^2 x$.
Therefore,$I = e^x \tan x + C$.