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.