Integrate the function: $e^{x} \left( \frac{1+\sin x}{1+\cos x} \right)$

We need to evaluate the integral $I = \int e^{x} \left( \frac{1+\sin x}{1+\cos x} \right) dx$.
Using trigonometric identities,$1+\sin x = \sin^{2} \frac{x}{2} + \cos^{2} \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} = (\sin \frac{x}{2} + \cos \frac{x}{2})^{2}$ and $1+\cos x = 2 \cos^{2} \frac{x}{2}$.
Substituting these into the expression:
$I = \int e^{x} \left( \frac{(\sin \frac{x}{2} + \cos \frac{x}{2})^{2}}{2 \cos^{2} \frac{x}{2}} \right) dx$
$I = \int e^{x} \cdot \frac{1}{2} \left( \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} + 1 \right)^{2} dx$
$I = \frac{1}{2} \int e^{x} (\tan \frac{x}{2} + 1)^{2} dx$
$I = \frac{1}{2} \int e^{x} (\tan^{2} \frac{x}{2} + 1 + 2 \tan \frac{x}{2}) dx$
Since $\tan^{2} \frac{x}{2} + 1 = \sec^{2} \frac{x}{2}$,we have:
$I = \int e^{x} (\frac{1}{2} \sec^{2} \frac{x}{2} + \tan \frac{x}{2}) dx$
Let $f(x) = \tan \frac{x}{2}$. Then $f'(x) = \frac{1}{2} \sec^{2} \frac{x}{2}$.
Using the standard integral formula $\int e^{x} (f(x) + f'(x)) dx = e^{x} f(x) + C$,we get:
$I = e^{x} \tan \frac{x}{2} + C$.