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To solve the given problem, we need to find the inverse Laplace transform of the function:

\[ \mathcal{L}^{-1} \left\{ \frac{1}{s^2 (s^2 + 3s - 1)} \right\} \]

Here’s a step-by-step solution to this problem:

1. **Factor the denominator if possible**: The denominator \(s^2 (s^2 + 3s - 1)\) does not easily factor into simpler components. We need to use partial fraction decomposition.

2. **Partial fraction decomposition**: We express the given fraction as a sum of simpler fractions. Let:

\[
\frac{1}{s^2 (s^2 + 3s - 1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs + D}{s^2 + 3s - 1}
\]

Multiplying through by the common denominator \(s^2 (s^2 + 3s - 1)\), we get:

\[
1 = A s (s^2 + 3s - 1) + B (s^2 + 3s - 1) + (Cs + D) s^2
\]

Expanding and collecting like terms:

\[
1 = (A + C)s^3 + (3A + B)s^2 + (A - B)s - B
\]

By equating coefficients of like powers of \(s\), we obtain:

- For \(s^3\): \(A + C = 0\)
- For \(s^2\): \(3A + B = 0\)
- For \(s\): \(A - B = 0\)
- Constant term: \(-B = 1\)

Solving these equations step by step:

- From \(-B = 1\), we get \(B = -1\).
- Substituting \(B = -1\) into \(A - B = 0\), we get \(A = -1\).
- Substituting \(A = -1\) and \(B = -1\) into \(3A + B = 0\), we get \(3(-1) + (-1) = -4 \neq 0\). There seems to be an inconsistency in the initial assumption of partial fractions. Let's re-evaluate the structure.

Due to complexity, it may be beneficial to refer to partial fraction techniques tailored for repeated and irreducible quadratic factors.

3. **Correct Partial Fraction Decomposition**:
   Since \(s^2 + 3s - 1\) has no simple roots, let's refine the decomposition form:

\[
\frac{1}{s^2 (s^2 + 3s - 1)} = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs + D}{s^2 + 3s - 1}
\]

Solve the coefficients \(A, B, C, D\):

\[
1 = (A + C)s^3 + (3A + B + D)s^2 + (A - B + 3C)s + (-D)
\]

Equate coefficients:
\[
A + C = 0
\]
\[
3A + B + D = 0
\]
\[
A - B + 3C = 0
\]
\[
-D = 1 \implies D = -1
\]

Then solving:
\[
3A + B - 1 = 0 \implies 3A + B = 1
\]
\[
A + 3C = 0
\]

Considering system:

By substitution:
\[
B = -3A - 1, \quad C = -A
\]
\[
A - (-3A - 1) - 3A = 0 \implies 5A + 1 = 0 \implies A = -\frac{1}{5}, \quad C = \frac{1}{5}, \quad B = \frac{2}{5}
\]

Partial fraction:
\[
\frac{1}{s^2 (s^2 + 3s - 1)} = -\frac{1}{5s} + \frac{2}{5s^2} + \frac{\frac{1}{5}s - 1}{s^2 + 3s - 1}
\]

4. **Inverse Laplace Transform**:

Using known transforms:
\[
\mathcal{L}^{-1}\left(\frac{A}{s}\right) = A
\]
\[
\mathcal{L}^{-1}\left(\frac{B}{s^2}\right) = Bt
\]
\[
\mathcal{L}^{-1}\left(\frac{Cs + D}{s^2 + 3s - 1}\right) = \text{use convolution and shifting}
\]

Result:
\[
-\frac{1}{5} + \frac{2}{5}t + \mathcal{L}^{-1}\left(\frac{\frac{1}{5}s - 1}{s^2 + 3s - 1}\right)
\]

Ask if user needs more details, also list more questions:

1. How to solve for Laplace Transform?
2. Solving quadratic equations?
3. How convolution integral is used in Laplace?
4. Can you solve other complex integrals?
5. What are special properties of Laplace transform?

Tip: Always verify coefficients properly in partial fraction.

Learn how to solve the inverse Laplace transform of a complex fraction decomposition involving partial fractions. This problem covers Laplace transform techniques, partial fraction decomposition, and convolution methods. Suitable for advanced undergraduate students.

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