We need to evaluate the definite integral:

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

Step-by-step solution:

1. Simplify the integrand: $\frac{e^{x+1}}{e^x} = \frac{e^x \cdot e^1}{e^x} = e^1 = e$

   So the integral simplifies to: $\int_0^1 e \, dx$

2. Now, we can factor out the constant $e$: $e \int_0^1 1 \, dx$

3. The integral of $1$ over the interval $[0, 1]$ is simply the length of the interval, which is $1 - 0 = 1$: $e \cdot 1 = e$

Thus, the value of the definite integral is:

$e$

Would you like more details on this? Here are 5 related questions you might find interesting:

1. How would the solution change if the limits of integration were different?
2. What is the interpretation of this integral geometrically?
3. How can you compute a similar integral for more complex functions?
4. How would you approach integrals involving trigonometric functions in the numerator?
5. Can we apply substitution or integration by parts to this problem?

Tip: Always simplify the integrand first to make the integral easier to solve!