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.