To evaluate the integral
\[
I = \int_{-1}^{1} \frac{dx}{e^x + 1},
\]
we can follow these steps:
### Step 1: Rewrite the Integral
We can rewrite the integrand by adding and subtracting \( e^x \):
\[
I = \int_{-1}^{1} \left( \frac{1 + e^x}{e^x + 1} - \frac{e^x}{e^x + 1} \right) dx.
\]
### Step 2: Split the Integral
Now, we can split the integral into two parts:
\[
I = \int_{-1}^{1} \frac{1 + e^x}{e^x + 1} dx - \int_{-1}^{1} \frac{e^x}{e^x + 1} dx.
\]
### Step 3: Simplify the First Integral
The first integral simplifies to:
\[
\int_{-1}^{1} 1 \, dx = 2.
\]
### Step 4: Evaluate the Second Integral
For the second integral, we can use the substitution \( t = e^x + 1 \). Then, we have:
\[
dt = e^x \, dx \quad \Rightarrow \quad dx = \frac{dt}{e^x} = \frac{dt}{t - 1}.
\]
When \( x = -1 \), \( t = e^{-1} + 1 = \frac{1}{e} + 1 \) and when \( x = 1 \), \( t = e^1 + 1 = e + 1 \).
Now substituting into the integral:
\[
\int_{-1}^{1} \frac{e^x}{e^x + 1} dx = \int_{\frac{1}{e} + 1}^{e + 1} \frac{t - 1}{t} \cdot \frac{dt}{t - 1} = \int_{\frac{1}{e} + 1}^{e + 1} \frac{1}{t} dt - \int_{\frac{1}{e} + 1}^{e + 1} dt.
\]
### Step 5: Evaluate the Logarithmic Integral
The first part evaluates to:
\[
\left[ \ln t \right]_{\frac{1}{e} + 1}^{e + 1} = \ln(e + 1) - \ln\left(\frac{1}{e} + 1\right).
\]
The second part evaluates to:
\[
\left[ t \right]_{\frac{1}{e} + 1}^{e + 1} = (e + 1) - \left(\frac{1}{e} + 1\right) = e - \frac{1}{e}.
\]
### Step 6: Combine the Results
Putting it all together, we have:
\[
I = 2 - \left( \ln(e + 1) - \ln\left(\frac{1}{e} + 1\right) + e - \frac{1}{e} \right).
\]
### Final Answer
Thus, the final result for the integral is:
\[
I = 2 - \left( \ln(e + 1) - \ln\left(\frac{1}{e} + 1\right) + e - \frac{1}{e} \right).
\]
