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please solve problem number 6 which great, insightful and step-by-step explanation!

The problem is asking to find the inverse Laplace transform of the function:

\[
F(s) = \frac{1}{s} \left(\frac{1}{s^2 + 1}\right)
\]

Let's solve this step-by-step.

### Step 1: Understand the given function

The function \( F(s) = \frac{1}{s} \cdot \frac{1}{s^2 + 1} \) can be written as:

\[
F(s) = \frac{1}{s(s^2 + 1)}
\]

We need to find the inverse Laplace transform of \( F(s) \), i.e., we want to find \( f(t) \) such that:

\[
f(t) = \mathcal{L}^{-1}\left\{F(s)\right\}
\]

### Step 2: Decompose the function using Partial Fraction Decomposition

To make the inverse Laplace transform easier, we'll use partial fraction decomposition on \( F(s) \).

Let’s express \( F(s) \) as:

\[
\frac{1}{s(s^2 + 1)} = \frac{A}{s} + \frac{Bs + C}{s^2 + 1}
\]

To find \( A \), \( B \), and \( C \), multiply both sides by \( s(s^2 + 1) \):

\[
1 = A(s^2 + 1) + (Bs + C)s
\]

This expands to:

\[
1 = A(s^2 + 1) + Bs^2 + Cs
\]

\[
1 = (A + B)s^2 + Cs + A
\]

Now, equate the coefficients of \( s^2 \), \( s \), and the constant term from both sides of the equation:

- Coefficient of \( s^2 \): \( A + B = 0 \)
- Coefficient of \( s \): \( C = 0 \)
- Constant term: \( A = 1 \)

### Step 3: Solve for \( A \), \( B \), and \( C \)

From \( A = 1 \) and \( A + B = 0 \), we find:

\[
A = 1, \quad B = -1, \quad C = 0
\]

So the partial fraction decomposition is:

\[
\frac{1}{s(s^2 + 1)} = \frac{1}{s} - \frac{s}{s^2 + 1}
\]

### Step 4: Find the inverse Laplace transforms

Now, take the inverse Laplace transform of each term separately.

1. **Inverse Laplace transform of \( \frac{1}{s} \):**

\[
\mathcal{L}^{-1}\left\{\frac{1}{s}\right\} = 1
\]

2. **Inverse Laplace transform of \( \frac{s}{s^2 + 1} \):**

\[
\mathcal{L}^{-1}\left\{\frac{s}{s^2 + 1}\right\} = \cos(t)
\]

### Step 5: Combine the results

Combine the results from the inverse Laplace transforms:

\[
f(t) = 1 - \cos(t)
\]

### Final Answer:

The inverse Laplace transform of \( F(s) = \frac{1}{s(s^2 + 1)} \) is:

\[
f(t) = 1 - \cos(t)
\]

This function \( f(t) = 1 - \cos(t) \) represents the solution in the time domain.

Would you like to see more details or have any questions?

Here are some related questions you might find helpful:
1. How can the convolution theorem be applied to find the inverse Laplace transform of products like \( \frac{1}{s(s^2 + 1)} \)?
2. What are some common Laplace transforms and their corresponding time-domain functions?
3. How does the inverse Laplace transform change if additional terms like \( e^{-as} \) are involved?
4. How can partial fraction decomposition be used in solving differential equations with Laplace transforms?
5. What are the applications of inverse Laplace transforms in solving real-world problems?

**Tip:** When working with Laplace transforms, keep a table of common transforms handy to quickly identify and apply the inverse transform.

Explore how to find the inverse Laplace transform of a complex function step-by-step. This problem involves understanding Laplace transforms, using partial fraction decomposition, and applying inverse Laplace transform formulas effectively.

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