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### Problem 2: Find the Inverse Laplace Transform of the Given Functions

#### 2a) \(\cot^{-1}(s + 1)\)

Finding the inverse Laplace transform of \(\cot^{-1}(s + 1)\) directly is not straightforward due to the lack of a simple inverse pair. This specific problem may require specialized methods or a lookup in a comprehensive table of Laplace transforms. Unfortunately, such a direct transform isn't readily available.

#### 2b) \(\frac{e^{-s}}{(s^2 + 4)^2}\)

To find the inverse Laplace transform of \(\frac{e^{-s}}{(s^2 + 4)^2}\), we utilize the second shifting theorem, which states:

\[
\mathcal{L}^{-1}\left\{e^{-as}F(s)\right\} = u(t-a)f(t-a)
\]

Here, \(a = 1\) and \(F(s) = \frac{1}{(s^2 + 4)^2}\).

First, we find the inverse Laplace transform of \(F(s) = \frac{1}{(s^2 + 4)^2}\). This is a known Laplace transform pair:

\[
\mathcal{L}^{-1}\left\{\frac{1}{(s^2 + 4)^2}\right\} = \frac{t \sin(2t)}{4}
\]

Using the second shifting theorem:

\[
\mathcal{L}^{-1}\left\{\frac{e^{-s}}{(s^2 + 4)^2}\right\} = u(t-1) \frac{(t-1) \sin(2(t-1))}{4}
\]

Therefore, the solution is:

\[
\mathcal{L}^{-1}\left\{\frac{e^{-s}}{(s^2 + 4)^2}\right\} = u(t-1) \frac{(t-1) \sin(2(t-1))}{4}
\]

#### 2c) \(\frac{s - 6}{s^3 - 2s^2 + 6s}\)

To find the inverse Laplace transform of \(\frac{s - 6}{s^3 - 2s^2 + 6s}\), we first factorize the denominator:

\[
s^3 - 2s^2 + 6s = s(s^2 - 2s + 6)
\]

Next, we perform partial fraction decomposition:

\[
\frac{s - 6}{s(s^2 - 2s + 6)} = \frac{A}{s} + \frac{Bs + C}{s^2 - 2s + 6}
\]

Multiplying both sides by the common denominator \(s(s^2 - 2s + 6)\), we get:

\[
s - 6 = A(s^2 - 2s + 6) + s(Bs + C)
\]

Expanding and equating coefficients:

\[
s - 6 = As^2 - 2As + 6A + Bs^2 + Cs
\]

\[
s - 6 = (A + B)s^2 + (C - 2A)s + 6A
\]

By equating coefficients, we get:

1. \(A + B = 0\)
2. \(C - 2A = 1\)
3. \(6A = -6\)

Solving these equations:

\[
A = -1, \quad B = 1, \quad C = 1
\]

Thus, the partial fractions are:

\[
\frac{s - 6}{s(s^2 - 2s + 6)} = \frac{-1}{s} + \frac{s + 1}{s^2 - 2s + 6}
\]

We can further decompose the second term:

\[
\frac{s + 1}{s^2 - 2s + 6} = \frac{s - 1 + 2}{(s-1)^2 + 5} = \frac{s-1}{(s-1)^2 + 5} + \frac{2}{(s-1)^2 + 5}
\]

The known Laplace transform pairs are:

\[
\mathcal{L}^{-1}\left\{\frac{s-1}{(s-1)^2 + 5}\right\} = e^t \cos(\sqrt{5}t)
\]

\[
\mathcal{L}^{-1}\left\{\frac{2}{(s-1)^2 + 5}\right\} = \frac{2}{\sqrt{5}} e^t \sin(\sqrt{5}t)
\]

Therefore, combining all parts, the inverse Laplace transform is:

\[
\mathcal{L}^{-1}\left\{\frac{s - 6}{s^3 - 2s^2 + 6s}\right\} = -1 + e^t \cos(\sqrt{5}t) + \frac{2}{\sqrt{5}} e^t \sin(\sqrt{5}t)
\]

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Would you like to proceed with the detailed solution for each part, or do you have any specific question or part you want to focus on?

Here are five related questions you might want to explore:

1. How to solve inverse Laplace transforms involving complex roots?
2. What are the applications of the second shifting theorem in Laplace transforms?
3. How to use partial fraction decomposition in Laplace transform problems?
4. How to handle Laplace transforms with polynomial numerators and denominators?
5. What are the common inverse Laplace transform pairs?

**Tip:** When solving inverse Laplace transforms, always check if you can simplify the problem using known Laplace pairs and properties.

Explore detailed solutions for finding the inverse Laplace transform of complex functions, including cotangent inverse and exponential functions. Learn about the Second Shifting Theorem and partial fraction decomposition techniques. Suitable for college-level students and beyond.

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