$\int \frac{e^x(1+\sin x)}{1+\cos x}\mathrm{d}x$ is equal to

$\begin{array}{l}\int \frac{e^x(1+\sin x)}{1+\cos x}\mathrm{d}x\\ =\int \frac{e^x[1+2\sin (x/2)\cos (x/2\left)\right]}{2{\cos }^2(x/2)}\mathrm{d}x\\ =\int e^x\left(\frac{1}{2}{\sec }^2\frac{x}{2}+\tan \frac{x}{2}\right)\mathrm{d}x\\ =\frac{1}{2}\int e^x{\sec }^2\frac{x}{2}\mathrm{d}x+\int e^x\tan \frac{x}{2}\mathrm{d}x\end{array}$

$\begin{array}{l}=\frac{1}{2}\int e^x{\sec }^2\frac{x}{2}\mathrm{d}x+e^x\tan \frac{x}{2}-\frac{1}{2}\int e^x{\sec }^2\frac{x}{2}+C\\ \left[\text{ integrating by parts }\right]\\ =e^x\tan \frac{x}{2}+C\end{array}$