To evaluate the integral \( \int e^{x} \frac{1 + \sin x}{1 + \cos x} \, dx \), we will follow a systematic approach.
### Step 1: Rewrite the Integral
We start with the integral:
\[
I = \int e^{x} \frac{1 + \sin x}{1 + \cos x} \, dx
\]
We can separate the fraction:
\[
I = \int e^{x} \left( \frac{1}{1 + \cos x} + \frac{\sin x}{1 + \cos x} \right) \, dx
\]
### Step 2: Simplify the Terms
Now, we can rewrite the second term. We know that:
\[
\sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2}
\]
and
\[
1 + \cos x = 2 \cos^2 \frac{x}{2}
\]
Thus, we can express the integral as:
\[
I = \int e^{x} \left( \frac{1}{1 + \cos x} + \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{1 + \cos x} \right) \, dx
\]
This simplifies to:
\[
I = \int e^{x} \left( \frac{1}{2 \cos^2 \frac{x}{2}} + \tan \frac{x}{2} \right) \, dx
\]
### Step 3: Split the Integral
Now, we can split the integral into two parts:
\[
I = \int e^{x} \frac{1}{2 \cos^2 \frac{x}{2}} \, dx + \int e^{x} \tan \frac{x}{2} \, dx
\]
### Step 4: Solve the First Integral
The first integral can be rewritten as:
\[
\int e^{x} \sec^2 \frac{x}{2} \, dx
\]
Using integration by parts, let:
- \( u = \sec^2 \frac{x}{2} \)
- \( dv = e^{x} \, dx \)
Then:
- \( du = \sec^2 \frac{x}{2} \tan \frac{x}{2} \cdot \frac{1}{2} \, dx \)
- \( v = e^{x} \)
Using integration by parts:
\[
\int u \, dv = uv - \int v \, du
\]
We get:
\[
\int e^{x} \sec^2 \frac{x}{2} \, dx = e^{x} \sec^2 \frac{x}{2} - \int e^{x} \cdot \frac{1}{2} \sec^2 \frac{x}{2} \tan \frac{x}{2} \, dx
\]
### Step 5: Solve the Second Integral
The second integral \( \int e^{x} \tan \frac{x}{2} \, dx \) can also be solved using integration by parts, where:
- \( u = \tan \frac{x}{2} \)
- \( dv = e^{x} \, dx \)
Following the same process as before, we will find the solution for this integral.
### Final Step: Combine Results
After solving both integrals, we combine the results to get:
\[
I = e^{x} \tan \frac{x}{2} + C
\]
### Final Answer
Thus, the evaluated integral is:
\[
\int e^{x} \frac{1 + \sin x}{1 + \cos x} \, dx = e^{x} \tan \frac{x}{2} + C
\]
