To find the inverse Laplace transform of the expression

$L^{-1} \left( \frac{(x+1)}{x+4} \cdot \frac{1}{x} \right),$

we first simplify the given expression:

$\frac{(x+1)}{x(x+4)}.$

Now, let's break this into partial fractions, so it becomes easier to find the inverse Laplace transform. We can decompose it as:

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

Step 1: Partial Fraction Decomposition

We want to find $A$ and $B$ such that:

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

Multiply both sides by $x(x+4)$ to eliminate the denominators:

$x+1 = A(x+4) + Bx.$

Simplify:

$x+1 = A(x) + 4A + B(x).$

This gives:

$x + 1 = (A + B)x + 4A.$

Now, equate coefficients of powers of $x$:

1. For $x$: $A + B = 1$,
2. Constant term: $4A = 1$.

Step 2: Solve for $A$ and $B$

From $4A = 1$, we get:

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

Substitute this into $A + B = 1$:

$\frac{1}{4} + B = 1 \quad \Rightarrow \quad B = 1 - \frac{1}{4} = \frac{3}{4}.$

Thus, the partial fraction decomposition is:

$\frac{x+1}{x(x+4)} = \frac{1}{4x} + \frac{3}{4(x+4)}.$

Step 3: Find the Inverse Laplace Transform

Now, we can easily find the inverse Laplace transform of each term:

1. $L^{-1} \left( \frac{1}{4x} \right)$: The inverse Laplace transform of $\frac{1}{x}$ is $1$, so this becomes: $\frac{1}{4} \cdot \delta(t),$ where $\delta(t)$ is the Dirac delta function.

2. $L^{-1} \left( \frac{3}{4(x+4)} \right)$: The inverse Laplace transform of $\frac{1}{x+a}$ is $e^{-at}$, so for $\frac{1}{x+4}$, the inverse transform is $e^{-4t}$. Therefore, this becomes: $\frac{3}{4} e^{-4t}.$

Final Answer:

The inverse Laplace transform of the given expression is:

$L^{-1} \left( \frac{x+1}{x(x+4)} \right) = \frac{1}{4} \delta(t) + \frac{3}{4} e^{-4t}.$

Let me know if you need further clarifications.

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Next Questions:

1. How do you use the Laplace transform to solve differential equations?
2. What is the significance of the Dirac delta function in Laplace transforms?
3. Can you explain how partial fractions help in simplifying Laplace transforms?
4. What are the conditions for a function to have a Laplace transform?
5. How does the inverse Laplace transform apply in control theory?

Tip: The Laplace transform is especially useful for solving linear differential equations with constant coefficients because it converts differential equations into algebraic equations.