It looks like you've uploaded an image of a mathematical expression that you want to solve using the inverse Laplace transform in partial fractions. Let's first break down the expression in the image:

The expression shown is: $\frac{1}{(s-2)(s+3)}$

Steps to solve using Partial Fraction Decomposition:

1. Express the function as a sum of simpler fractions: $\frac{1}{(s-2)(s+3)} = \frac{A}{s-2} + \frac{B}{s+3}$ where $A$ and $B$ are constants to be determined.

2. Multiply both sides by $(s-2)(s+3)$ to get rid of the denominator: $1 = A(s + 3) + B(s - 2)$ Simplify and group like terms: $1 = (A + B)s + (3A - 2B)$

3. Set up a system of equations by equating coefficients:

   • Coefficient of $s$: $A + B = 0$
   • Constant term: $3A - 2B = 1$

4. Solve the system of equations:

   • From $A + B = 0$, we have $B = -A$.
   • Substitute into the second equation: $3A - 2(-A) = 1$, which simplifies to $3A + 2A = 1$, so $5A = 1$, hence $A = \frac{1}{5}$.
   • Since $B = -A$, $B = -\frac{1}{5}$.

   So, the partial fraction decomposition is: $\frac{1}{(s-2)(s+3)} = \frac{\frac{1}{5}}{s-2} - \frac{\frac{1}{5}}{s+3}$

5. Apply the inverse Laplace transform:

   • The inverse Laplace transform of $\frac{1}{s-a}$ is $e^{at}$.
   • Therefore: $\mathcal{L}^{-1}\left(\frac{1}{(s-2)(s+3)}\right) = \frac{1}{5}e^{2t} - \frac{1}{5}e^{-3t}$

Final Answer:

$\mathcal{L}^{-1}\left(\frac{1}{(s-2)(s+3)}\right) = \frac{1}{5}e^{2t} - \frac{1}{5}e^{-3t}$

Would you like any further details or have any questions on this solution?

Here are some related questions you might consider:

1. How does the Laplace transform of $\frac{1}{s^2 + a^2}$ compare to this case?
2. What happens if the denominator has repeated roots?
3. How can partial fraction decomposition be applied to higher-degree polynomials?
4. What is the significance of each term in the inverse Laplace transform?
5. How would the procedure change if $s$ had a complex root?

Tip: When solving inverse Laplace problems, always make sure your decomposition is correct by re-multiplying to check your work before applying the inverse transform.