Let's find the Laplace transform of the given function:

$f(t) = \int_0^{t} e^u \cosh(u) \, du$

Step 1: Solve the integral

The integrand is $e^u \cosh(u)$, which simplifies as follows:

Recall that $\cosh(u) = \frac{e^u + e^{-u}}{2}$, so:

$e^u \cosh(u) = e^u \left( \frac{e^u + e^{-u}}{2} \right) = \frac{e^{2u} + 1}{2}$

Thus, the integral becomes:

$f(t) = \int_0^{t} \frac{e^{2u} + 1}{2} \, du$

Now, let's evaluate the integral term-by-term:

$\int_0^{t} \frac{e^{2u}}{2} \, du = \frac{1}{2} \cdot \frac{e^{2u}}{2} \bigg|_0^t = \frac{e^{2t} - 1}{4}$

$\int_0^{t} \frac{1}{2} \, du = \frac{t}{2}$

Thus, the result of the integral is:

$f(t) = \frac{e^{2t} - 1}{4} + \frac{t}{2}$

Step 2: Find the Laplace transform of $f(t)$

Now, we take the Laplace transform of $f(t) = \frac{e^{2t} - 1}{4} + \frac{t}{2}$.

Laplace transform of $\frac{e^{2t} - 1}{4}$:

The Laplace transform of $e^{at}$ is $\frac{1}{s - a}$, so:

$\mathcal{L}\left\{ e^{2t} \right\} = \frac{1}{s - 2}$

The Laplace transform of a constant (in this case, -1) is $\frac{1}{s}$, so:

$\mathcal{L}\left\{ \frac{e^{2t} - 1}{4} \right\} = \frac{1}{4} \left( \frac{1}{s - 2} - \frac{1}{s} \right)$

Laplace transform of $\frac{t}{2}$:

The Laplace transform of $t$ is $\frac{1}{s^2}$, so:

$\mathcal{L}\left\{ \frac{t}{2} \right\} = \frac{1}{2s^2}$

Step 3: Combine the results

Now, combine all the Laplace transforms:

$\mathcal{L}\{ f(t) \} = \frac{1}{4} \left( \frac{1}{s - 2} - \frac{1}{s} \right) + \frac{1}{2s^2}$

Thus, the Laplace transform of $\int_0^{t} e^u \cosh(u) \, du$ is:

$\boxed{\mathcal{L}\{ f(t) \} = \frac{1}{4} \left( \frac{1}{s - 2} - \frac{1}{s} \right) + \frac{1}{2s^2}}$

Would you like more details or have any further questions?

Here are 5 related questions:

1. What is the Laplace transform of $\cosh(t)$?
2. How would you find the Laplace transform of $e^t \sin(t)$?
3. What is the Laplace transform of $t^n$ for a general $n$?
4. How do you compute the inverse Laplace transform of a rational function?
5. What are the basic properties of the Laplace transform?

Tip: The linearity of the Laplace transform makes it much easier to break down complex functions into simpler terms to transform individually!