We are tasked with finding the inverse Laplace transform of the expression:

$\mathcal{L}^{-1} \left\{ \frac{(p+1)}{p(p+4)} \right\}$

To solve this, we can use partial fraction decomposition to simplify the expression into terms that have known inverse Laplace transforms.

Step 1: Partial fraction decomposition

We express $\frac{(p+1)}{p(p+4)}$ as:

$\frac{(p+1)}{p(p+4)} = \frac{A}{p} + \frac{B}{p+4}$

Now, solve for $A$ and $B$:

$(p+1) = A(p+4) + Bp$

Expanding the right-hand side:

$p+1 = A(p+4) + Bp = A(p) + 4A + Bp$

Grouping like terms:

$p+1 = (A+B)p + 4A$

Equating the coefficients of $p$ and the constant terms:

1. $A + B = 1$
2. $4A = 1$

From equation 2:

$A = \frac{1}{4}$

Substitute $A = \frac{1}{4}$ into equation 1:

$\frac{1}{4} + B = 1$

Solving for $B$:

$B = 1 - \frac{1}{4} = \frac{3}{4}$

Step 2: Rewrite the expression

Now we can rewrite the original expression as:

$\frac{(p+1)}{p(p+4)} = \frac{1}{4} \cdot \frac{1}{p} + \frac{3}{4} \cdot \frac{1}{p+4}$

Step 3: Apply the inverse Laplace transform

Using known inverse Laplace transforms:

1. $\mathcal{L}^{-1} \left\{ \frac{1}{p} \right\} = 1$
2. $\mathcal{L}^{-1} \left\{ \frac{1}{p+a} \right\} = e^{-at}$

Thus:

$\mathcal{L}^{-1} \left\{ \frac{1}{4} \cdot \frac{1}{p} + \frac{3}{4} \cdot \frac{1}{p+4} \right\} = \frac{1}{4} \cdot 1 + \frac{3}{4} \cdot e^{-4t}$

$\mathcal{L}^{-1} \left\{ \frac{(p+1)}{p(p+4)} \right\} = \frac{1}{4} + \frac{3}{4} e^{-4t}$

Final Answer:

$\mathcal{L}^{-1} \left\{ \frac{(p+1)}{p(p+4)} \right\} = \frac{1}{4} + \frac{3}{4} e^{-4t}$

Would you like more details on any step? Here are five related questions to expand on this:

1. How does partial fraction decomposition work for rational functions?
2. What are the general properties of the inverse Laplace transform?
3. How can we solve inverse Laplace problems involving more complex numerators?
4. What are some common inverse Laplace transforms?
5. How is the inverse Laplace transform applied in solving differential equations?

Tip: When performing partial fraction decomposition, ensure the degree of the numerator is less than the denominator before proceeding!