Show that

$\begin{aligned} I &=\int\left(\frac{2+\sin 2 x}{1+\cos 2 x}\right) e^{x} \\ &=\int\left(\frac{2+2 \sin x \cos x}{2 \cos ^{2} x}\right) e^{x} \\ &=\int\left(\frac{1+\sin x \cos x}{\cos ^{2} x}\right) e^{x} \\ &=\int\left(\sec ^{2} x+\tan x\right) e^{x} \end{aligned}$

Let $f(x)=\tan x \Rightarrow f^{\prime}(x)=\sec ^{2} x$

$\therefore I=\int\left(f(x)+f^{\prime}(x)\right] e^{x} d x$

$=e^{x} f(x)+\mathrm{C}$

$=e^{x} \tan x+\mathrm{C}$