To solve the problem, we need to find the Laplace transform $L\left\{\int_0^t u \sinh(2u) \, du \right\}$.

Step-by-Step Approach:

1. Recall the property of Laplace Transform: The Laplace transform of an integral is given by: $L\left\{\int_0^t f(u) \, du \right\} = \frac{F(s)}{s}$ where $F(s) = L\{f(t)\}$ is the Laplace transform of $f(t)$.

2. Find the Laplace transform of $u \sinh(2u)$: Let's compute the Laplace transform of the function $f(u) = u \sinh(2u)$.

   • The Laplace transform of $\sinh(2u)$ is: $L\{\sinh(2u)\} = \frac{2}{s^2 - 4}$

   • Now, to find $L\{u \sinh(2u)\}$, we use the following known result for $t f(t)$: $L\{t f(t)\} = -\frac{d}{ds} L\{f(t)\}$ Therefore, applying this to $f(u) = \sinh(2u)$: $L\{u \sinh(2u)\} = -\frac{d}{ds} \left(\frac{2}{s^2 - 4}\right)$

3. Differentiate the expression: Let's differentiate $\frac{2}{s^2 - 4}$ with respect to $s$: $\frac{d}{ds} \left( \frac{2}{s^2 - 4} \right) = 2 \cdot \frac{-2s}{(s^2 - 4)^2} = \frac{-4s}{(s^2 - 4)^2}$

4. Apply the integral rule: Using the Laplace transform property of integrals mentioned earlier: $L\left\{ \int_0^t u \sinh(2u) \, du \right\} = \frac{L\{u \sinh(2u)\}}{s} = \frac{-4s}{s(s^2 - 4)^2}$

   Simplifying: $\frac{-4s}{s(s^2 - 4)^2} = \frac{-4}{(s^2 - 4)^2}$

Conclusion:

The correct answer is option (a): $\frac{-4}{(s^2 - 4)^2}$

Do you want more details on any step or have additional questions? Here are five related questions:

1. What is the general process for finding Laplace transforms of integrals?
2. How can we differentiate expressions involving complex fractions?
3. What are the key properties of the hyperbolic sine function, $\sinh(x)$?
4. How does the differentiation property of Laplace transforms work for various functions?
5. Can the Laplace transform be used to solve differential equations involving integrals?

Tip: The Laplace transform simplifies complex integral and differential operations, making it a powerful tool for solving linear systems.