We are asked to find the inverse Laplace transform of the function:

$F(s) = \frac{3s^2 + 1}{s^2(s^2 + 1)}$

Step 1: Decompose the function

We start by breaking down $F(s)$ into simpler fractions using partial fraction decomposition.

$F(s) = \frac{3s^2 + 1}{s^2(s^2 + 1)}$

Let’s rewrite this as:

$F(s) = \frac{A}{s} + \frac{B}{s^2} + \frac{Cs + D}{s^2 + 1}$

Multiplying both sides by $s^2(s^2 + 1)$ to clear the denominators:

$3s^2 + 1 = A s(s^2 + 1) + B(s^2 + 1) + (Cs + D)s^2$

Expanding the right-hand side:

$3s^2 + 1 = A s^3 + A s + B s^2 + B + C s^3 + D s^2$

Group terms by powers of $s$:

$3s^2 + 1 = (A + C) s^3 + (B + D) s^2 + A s + B$

Step 2: Match coefficients

We can now match the coefficients of $s^3$, $s^2$, $s$, and the constant:

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

From $A = 0$, $C = 0$, $B = 1$, and $D = 3$, the partial fraction decomposition becomes:

$F(s) = \frac{1}{s^2} + \frac{3}{s^2 + 1}$

Step 3: Find the inverse Laplace transforms

• The inverse Laplace transform of $\frac{1}{s^2}$ is $t$.
• The inverse Laplace transform of $\frac{3}{s^2 + 1}$ is $3 \sin(t)$.

Step 4: Combine the results

The inverse Laplace transform of $F(s)$ is:

$f(t) = t + 3 \sin(t)$

Step 5: Match with the given options

This matches option (a): $2 \sin(t) + t$.

Final Answer:

The correct answer is (a).

Would you like further details or explanations? Here are some related questions to expand your understanding:

1. How is partial fraction decomposition useful in inverse Laplace transforms?
2. What is the inverse Laplace transform of $\frac{1}{s(s^2 + 1)}$?
3. How does Laplace transform simplify solving differential equations?
4. How can you identify when to use inverse Laplace transforms in a problem?
5. What is the general approach to solving Laplace transforms involving polynomials?

Tip: Always ensure you match terms carefully when using partial fractions to avoid errors in coefficients!