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The image contains two integrals involving functions \( u_1(t) \) and \( u_2(t) \), both with respect to \( t \). The integrals are as follows:

\[
u_1(t) = \int \frac{2}{e^{3t}(2 + e^t)} \, dt
\]
\[
u_2(t) = \int \frac{-2}{2 + e^t} \, dt
\]

### Let's solve each integral step by step.

#### 1. **For \( u_1(t) \):**

\[
u_1(t) = \int \frac{2}{e^{3t}(2 + e^t)} \, dt
\]

First, factor the exponential function:

\[
u_1(t) = \int \frac{2}{e^{3t}(2 + e^t)} \, dt = \int \frac{2}{e^{2t}(2e^{-t} + 1)} \, dt
\]

Let \( x = e^t \), which implies \( dx = e^t dt \) and \( dt = \frac{dx}{x} \). Rewriting the integral in terms of \( x \):

\[
u_1(t) = \int \frac{2}{e^{2t}(2 + x)} \, dt
\]

I will complete the substitution process for this integral.The solution to the first integral, \( u_1(t) \), is:

\[
u_1(t) = \frac{3(-t + \log(e^t + 2))e^{3t} - 6e^{2t} + 6e^t - 8}{24e^{3t}}
\]

This can be further simplified if necessary, but the current form is correct.

#### 2. **For \( u_2(t) \):**

\[
u_2(t) = \int \frac{-2}{2 + e^t} \, dt
\]

For this integral, let \( x = e^t \), so \( dx = e^t dt \) and \( dt = \frac{dx}{x} \). The integral becomes:

\[
u_2(t) = \int \frac{-2}{2 + x} \cdot \frac{dx}{x}
\]

I will now solve this integral.The solution to the second integral, \( u_2(t) \), is:

\[
u_2(t) = -t + \log(e^t + 2)
\]

### Summary of the results:

1. \( u_1(t) = \frac{3(-t + \log(e^t + 2))e^{3t} - 6e^{2t} + 6e^t - 8}{24e^{3t}} \)
2. \( u_2(t) = -t + \log(e^t + 2) \)

Would you like any further details or clarification on these solutions?

Here are 5 related questions for further exploration:

1. How can you simplify the expression for \( u_1(t) \)?
2. What substitution methods can be used to solve more complex integrals involving exponentials?
3. How does \( u_2(t) \) behave as \( t \to \infty \)?
4. Can you express \( u_1(t) \) in terms of a Taylor series?
5. What are the applications of integrals involving exponentials in physics?

**Tip**: When solving integrals, substitutions often simplify complicated expressions, especially when dealing with exponential functions.

Solve the following integrals: u1(t) = ∫(2 / (e^(3t) * (2 + e^t))) dt and u2(t) = ∫(-2 / (2 + e^t)) dt.

Explore the step-by-step solution to two integrals involving exponential functions: u1(t) = ∫(2 / (e^(3t) * (2 + e^t))) dt and u2(t) = ∫(-2 / (2 + e^t)) dt. This solution applies substitution techniques and is ideal for calculus students at an undergraduate level.

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