The integral $\mathrm{I}=\int e^x\left(\frac{1+\sin x}{1+\cos x}\right) d x$ is :

$e^x \tan \left(\frac{x}{2}\right)+C$, where C is a constant

$I=\int e^x\left(\frac{1+\sin x}{1+\cos x}\right) d x$

$\Rightarrow I=\int e^x\left[\frac{\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}\right] d x$

as $sin^2 \frac{x}{2} + cos^2 \frac{x}{2} = 1$

$sin x = 2 sin \frac{x}{2} cos \frac{x}{2}$

$1 + cos^x = 2 cos^2 \frac{x}{2}$

$=\int e^x\left[\frac{1^2}{2 \cos ^2 \frac{x}{2}}+\frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \cos ^2 \frac{x}{2}}\right] d x$

$=\int e^x\left[\frac{1}{2} \sec ^2 \frac{x}{2}+\tan \frac{x}{2}\right] d x$

this of the form

$\int e^x (f'(x) + f(x)) dx = e^x f(x) + C$

$f(x) = tan \frac{x}{2}  f'(x) = \frac{1}{2} sec^2 \frac{x}{2}$

$e^x tan \frac{x}{2} + C$