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To solve the integral 

\[
\int_0^1 \frac{e^x + 1}{e^x} \, dx,
\]

let's simplify the integrand before integrating.

### Step 1: Simplify the Integrand
We can rewrite the fraction:

\[
\frac{e^x + 1}{e^x} = \frac{e^x}{e^x} + \frac{1}{e^x} = 1 + e^{-x}.
\]

Thus, the integral becomes:

\[
\int_0^1 \left(1 + e^{-x}\right) \, dx.
\]

### Step 2: Separate the Integral
Now we can split this into two simpler integrals:

\[
\int_0^1 \left(1 + e^{-x}\right) \, dx = \int_0^1 1 \, dx + \int_0^1 e^{-x} \, dx.
\]

### Step 3: Evaluate Each Integral
1. **First Integral**: \(\int_0^1 1 \, dx\)

   \[
   \int_0^1 1 \, dx = \left[ x \right]_0^1 = 1 - 0 = 1.
   \]

2. **Second Integral**: \(\int_0^1 e^{-x} \, dx\)

   The antiderivative of \(e^{-x}\) is \(-e^{-x}\). Thus,

   \[
   \int_0^1 e^{-x} \, dx = \left[ -e^{-x} \right]_0^1 = -e^{-1} + e^0 = -\frac{1}{e} + 1.
   \]

### Step 4: Combine Results
Now, adding the two results together:

\[
\int_0^1 \frac{e^x + 1}{e^x} \, dx = 1 + \left(-\frac{1}{e} + 1\right) = 1 + 1 - \frac{1}{e} = 2 - \frac{1}{e}.
\]

### Final Answer
\[
\int_0^1 \frac{e^x + 1}{e^x} \, dx = 2 - \frac{1}{e}.
\]

---

Would you like further details on any step?

Here are some additional related questions:
1. What is the integral of \( e^{-x} \) over different limits?
2. How can substitution be used for integrals of exponential functions?
3. What is the importance of simplifying the integrand before integrating?
4. Can this problem be approached with a different method?
5. How does \( e^{-x} \) behave as \( x \to \infty \)?

**Tip**: Simplifying the integrand before starting the integration often makes the process easier and reduces errors.

Calculate \( \int_0^1 \frac{e^x + 1}{e^x} \, dx \)

Step-by-step solution for calculating the integral \( \int_0^1 \frac{e^x + 1}{e^x} \, dx \). This solution covers simplifying the integrand, breaking it into simpler parts, and evaluating each integral using basic exponential rules. Suitable for undergraduate calculus students.

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