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This image presents a problem involving the inverse Laplace transform \( L^{-1} \) applied to the function:

\[
L^{-1}\left\{\frac{s}{(s^2 + 9)(s^2 + 16)}\right\}
\]

The options are expressions involving trigonometric functions \( \cos(4t) \) and \( \cos(3t) \). To solve this, we need to decompose the given rational function using partial fraction decomposition and apply the inverse Laplace transform to each term.

### Step-by-Step Solution:

1. **Decompose the given function**:
   \[
   \frac{s}{(s^2 + 9)(s^2 + 16)}
   \]
   We will express this as a sum of partial fractions:
   \[
   \frac{s}{(s^2 + 9)(s^2 + 16)} = \frac{A s}{s^2 + 9} + \frac{B s}{s^2 + 16}
   \]
   where \( A \) and \( B \) are constants to be determined.

2. **Multiply both sides by \( (s^2 + 9)(s^2 + 16) \)** to eliminate the denominators:
   \[
   s = A s(s^2 + 16) + B s(s^2 + 9)
   \]
   Expand both terms:
   \[
   s = A s(s^2 + 16) + B s(s^2 + 9) = A s^3 + 16A s + B s^3 + 9B s
   \]
   Group terms involving \( s \):
   \[
   s = (A + B) s^3 + (16A + 9B) s
   \]
   For this to be true for all \( s \), the coefficients of the powers of \( s \) must match. So we get the system of equations:
   - \( A + B = 0 \) (from the \( s^3 \) term),
   - \( 16A + 9B = 1 \) (from the \( s \) term).

3. **Solve for \( A \) and \( B \)**:
   From \( A + B = 0 \), we have \( B = -A \). Substitute this into the second equation:
   \[
   16A + 9(-A) = 1 \implies 16A - 9A = 1 \implies 7A = 1 \implies A = \frac{1}{7}
   \]
   Therefore, \( B = -\frac{1}{7} \).

4. **Substitute \( A \) and \( B \) into the partial fraction decomposition**:
   \[
   \frac{s}{(s^2 + 9)(s^2 + 16)} = \frac{1/7 \cdot s}{s^2 + 9} + \frac{-1/7 \cdot s}{s^2 + 16}
   \]

5. **Apply the inverse Laplace transform**:
   - The inverse Laplace transform of \( \frac{s}{s^2 + a^2} \) is \( \cos(at) \).
   - Thus,
     \[
     L^{-1}\left\{ \frac{1/7 \cdot s}{s^2 + 9} \right\} = \frac{1}{7} \cos(3t)
     \]
     and
     \[
     L^{-1}\left\{ \frac{-1/7 \cdot s}{s^2 + 16} \right\} = -\frac{1}{7} \cos(4t)
     \]

6. **Final result**:
   \[
   L^{-1}\left\{\frac{s}{(s^2 + 9)(s^2 + 16)}\right\} = \frac{1}{7} \cos(3t) - \frac{1}{7} \cos(4t)
   \]

This matches option **(d)**.

### Answer: (d)

Let me know if you would like further details or have any questions!

### Related Questions:
1. What is the general form for the inverse Laplace transform of functions involving trigonometric terms?
2. How do you handle inverse Laplace transforms of second-order polynomials?
3. Can the inverse Laplace transform method be applied to differential equations directly?
4. How do you approach partial fraction decomposition when the roots are complex?
5. What is the relation between Laplace transforms and Fourier transforms in solving differential equations?

### Tip:
For Laplace transforms of trigonometric functions, memorize the standard forms such as \( \frac{s}{s^2 + a^2} \) for cosine and \( \frac{a}{s^2 + a^2} \) for sine functions. This helps in solving inverse transforms quickly.

What is the inverse Laplace transform of the function \( L^{-1}\left\{\frac{s}{(s^2 + 9)(s^2 + 16)}\right\} \)? The options include trigonometric functions like \( \cos(3t) \) and \( \cos(4t) \).

Learn how to find the inverse Laplace transform of \( \frac{s}{(s^2 + 9)(s^2 + 16)} \) using partial fraction decomposition and trigonometric identities. This problem is ideal for students studying differential equations at the undergraduate level.

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