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We are tasked with finding the Laplace transform of the integral:

\[
L \left\{ \int_0^t u \cosh(2u) \, du \right\}
\]

To approach this, let's break it down step-by-step:

1. **Laplace Transform of the Integral**:
   We know from the properties of Laplace transforms that:
   \[
   L \left\{ \int_0^t f(u) \, du \right\} = \frac{F(s)}{s}
   \]
   where \( F(s) = L\{ f(u) \} \).

2. **Function to Transform**:
   The function inside the integral is \( u \cosh(2u) \). We need the Laplace transform of \( u \cosh(2u) \).

3. **Laplace Transform of \( \cosh(2u) \)**:
   The Laplace transform of \( \cosh(au) \) is:
   \[
   L\{ \cosh(au) \} = \frac{s}{s^2 - a^2}
   \]
   For \( a = 2 \), this becomes:
   \[
   L\{ \cosh(2u) \} = \frac{s}{s^2 - 4}
   \]

4. **Laplace Transform of \( u \cosh(2u) \)**:
   To find \( L\{ u \cosh(2u) \} \), we use the formula for the Laplace transform of \( u f(u) \):
   \[
   L\{ u f(u) \} = -\frac{d}{ds} L\{ f(u) \}
   \]
   Applying this to \( f(u) = \cosh(2u) \), we differentiate:
   \[
   L\{ u \cosh(2u) \} = -\frac{d}{ds} \left( \frac{s}{s^2 - 4} \right)
   \]

5. **Differentiation**:
   Differentiate \( \frac{s}{s^2 - 4} \) with respect to \( s \):
   \[
   \frac{d}{ds} \left( \frac{s}{s^2 - 4} \right) = \frac{(s^2 - 4) - s(2s)}{(s^2 - 4)^2} = \frac{s^2 - 4 - 2s^2}{(s^2 - 4)^2} = \frac{-s^2 - 4}{(s^2 - 4)^2}
   \]
   So:
   \[
   L\{ u \cosh(2u) \} = \frac{-(s^2 + 4)}{(s^2 - 4)^2}
   \]

6. **Final Step**:
   Now, applying the Laplace transform property for integrals:
   \[
   L\left\{ \int_0^t u \cosh(2u) \, du \right\} = \frac{L\{ u \cosh(2u) \}}{s} = \frac{-(s^2 + 4)}{s(s^2 - 4)^2}
   \]

Thus, the correct answer is:

\[
\boxed{d}
\]

Would you like a deeper explanation on any of these steps?

Here are some related questions for further exploration:
1. How do you compute the Laplace transform of other hyperbolic functions?
2. What are the properties of the Laplace transform that involve differentiation?
3. How would you approach this problem if it involved sine or cosine instead of cosh?
4. Can you generalize this technique for Laplace transforms of other integrals?
5. How do you find the inverse Laplace transform of similar expressions?

**Tip:** Always remember to utilize properties of Laplace transforms like differentiation and integration to simplify complex expressions efficiently.

Find the Laplace transform L{∫_0^t u cosh(2u) du}

Learn how to compute the Laplace transform of ∫_0^t u cosh(2u) du, using integral transform properties and hyperbolic functions. This detailed guide walks through the step-by-step process, applying Laplace transformation rules and differentiation.

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