https://file.aiquickdraw.com/mathbot/file/1725948623135jj1pe12p.jpg

The given integral in the image is:

\[
\int_0^{\ln(2)} \frac{\ln(1 + e^x)}{e^x} \, dx
\]

Let me proceed by solving this step by step.

### Step 1: Substitution
We can simplify the integrand by substituting \( e^x \). Let:
\[
 u = e^x \implies du = e^x dx
\]
When \( x = 0 \), \( u = 1 \). When \( x = \ln(2) \), \( u = 2 \).

Now the integral becomes:
\[
 \int_1^2 \frac{\ln(1 + u)}{u} \, du
\]

### Step 2: Recognizing the integral
This is a standard integral form:
\[
 \int \frac{\ln(1+u)}{u} \, du
\]

It is known that:
\[
 \int \frac{\ln(1+u)}{u} \, du = \frac{1}{2} \ln^2(1 + u)
\]

### Step 3: Applying limits
Now apply the limits from \( u = 1 \) to \( u = 2 \):
\[
 \frac{1}{2} \left[ \ln^2(1 + 2) - \ln^2(1 + 1) \right]
 = \frac{1}{2} \left[ \ln^2(3) - \ln^2(2) \right]
\]

Thus, the value of the integral is:
\[
 \frac{1}{2} \left( \ln^2(3) - \ln^2(2) \right)
\]

This is the final answer.

### Final Result:
\[
 \int_0^{\ln(2)} \frac{\ln(1 + e^x)}{e^x} \, dx = \frac{1}{2} \left( \ln^2(3) - \ln^2(2) \right)
\]

Would you like any further details or clarifications?

Here are 5 related questions you might want to explore:
1. How can you solve integrals involving logarithmic and exponential functions?
2. What is the significance of substitution in solving integrals?
3. How would you handle integrals with more complex limits of integration?
4. Can you generalize this type of integral for different limits?
5. How would you approach solving this integral using numerical methods?

**Tip**: When faced with integrals involving logarithmic functions, always look for opportunities to use substitution to simplify the terms.

Solve the integral \( \int_0^{\ln(2)} \frac{\ln(1 + e^x)}{e^x} \, dx \)

Learn how to solve the integral \( \int_0^{\ln(2)} \frac{\ln(1 + e^x)}{e^x} \, dx \) step-by-step. This problem involves logarithmic and exponential functions, substitution techniques, and applies the formula for \( \int \frac{\ln(1+u)}{u} \, du \). Suitable for advanced high school and university-level students.

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