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.