Find the inverse Laplace transform of the following: 1. $$ F(s)=\frac{1}{s^{2}+3 s+2} $$ 2. $$ F(s)=\frac{s+2}{s\left(s^{2}+9 s+20\right)} $$ 3. $$ F(s)=\frac{2 s+4}{s^{3}+6 s^{2}+11 s+6} $$ 4. $$ F(s)=\frac{s+3}{s^{2}+10 s+34} $$ 5. $$ F(s)=\frac{s+3}{s^{2}(s+1)} $$

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To find the inverse Laplace transforms of the given functions, we will first factor the denominators where necessary and then apply the appropriate techniques for inverse Laplace transforms.

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

First, we factor the denominator:
$$
s^{2}+3 s+2 = (s+1)(s+2)
$$
Thus, we can rewrite $$ F(s) $$:
$$
F(s) = \frac{1}{(s+1)(s+2)}
$$

Next, we can use partial fraction decomposition:
$$
\frac{1}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2}
$$

Multiplying through by the denominator:
$$
1 = A(s+2) + B(s+1)
$$

Setting $$ s = -1 $$:
$$
1 = A(1) \implies A = 1
$$

Setting $$ s = -2 $$:
$$
1 = B(-1) \implies B = -1
$$

Thus, we have:
$$
F(s) = \frac{1}{s+1} - \frac{1}{s+2}
$$

The inverse Laplace transform is:
$$
\mathcal{L}^{-1}\{F(s)\} = e^{-t} - e^{-2t}
$$

### 2. $$ F(s)=\frac{s+2}{s\left(s^{2}+9 s+20\right)} $$

First, we factor the quadratic:
$$
s^{2}+9 s+20 = (s+4)(s+5)
$$
Thus, we can rewrite $$ F(s) $$:
$$
F(s) = \frac{s+2}{s(s+4)(s+5)}
$$

Using partial fraction decomposition:
$$
\frac{s+2}{s(s+4)(s+5)} = \frac{A}{s} + \frac{B}{s+4} + \frac{C}{s+5}
$$

Multiplying through by the denominator:
$$
s+2 = A(s+4)(s+5) + B(s)(s+5) + C(s)(s+4)
$$

Setting $$ s = 0 $$:
$$
2 = A(20) \implies A = \frac{1}{10}
$$

Setting $$ s = -4 $$:
$$
-2 = B(-4)(1) \implies B = \frac{1}{2}
$$

Setting $$ s = -5 $$:
$$
-3 = C(-5)(-1) \implies C = \frac{3}{5}
$$

Thus, we have:
$$
F(s) = \frac{1/10}{s} + \frac{1/2}{s+4} + \frac{3/5}{s+5}
$$

The inverse Laplace transform is:
$$
\mathcal{L}^{-1}\{F(s)\} = \frac{1}{10} + \frac{1}{2}e^{-4t} + \frac{3}{5}e^{-5t}
$$

### 3. $$ F(s)=\frac{2 s+4}{s^{3}+6 s^{2}+11 s+6} $$

First, we factor the cubic polynomial:
$$
s^{3}+6 s^{2}+11 s+6 = (s+1)(s+2)(s+3)
$$
Thus, we can rewrite $$ F(s) $$:
$$
F(s) = \frac{2 s+4}{(s+1)(s+2)(s+3)}
$$

Using partial fraction decomposition:
$$
\frac{2 s+4}{(s+1)(s+2)(s+3)} = \frac{A}{s+1} + \frac{B}{s+2} + \frac{C}{s+3}
$$

Multiplying through by the denominator:
$$
2s + 4 = A(s+2)(s+3) + B(s+1)(s+3) + C(s+1)(s+2)
$$

Setting $$ s = -1 $$:
$$
2(-1) + 4 = A(1)(2) \implies 2 = 2A \implies A = 1
$$

Setting $$ s = -2 $$:
$$
2(-2) + 4 = B(-1)(1) \implies 0 = -B \implies B = 0
$$

Setting $$ s = -3 $$:
$$
2(-3) + 4 = C(-2)(-1) \implies -2 = 2C \implies C = -1
$$

Thus, we have:
$$
F(s) = \frac{1}{s+1} - \frac{1}{s+3}
$$

The inverse Laplace transform is:
$$
\mathcal{L}^{-1}\{F(s)\} = e^{-t} - e^{-3t}
$$

### 4. $$ F(s)=\frac{s+3}{s^{2}+10 s+34} $$

First, we complete the square for the quadratic:
$$
s^{2}+10 s+34 = (s+5)^{2} + 9
$$
Thus, we can rewrite $$ F(s) $$:
$$
F(s) = \frac{s+3}{(s+5)^{2}+3^{2}}
$$

The inverse Laplace transform is:
$$
\mathcal{L}^{-1}\{F(s)\} = e^{-5t}\left(\cos(3t) + \frac{1}{3}\sin(3t)\right)
$$

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

Using partial fraction decomposition:
$$
\frac{s+3}{s^{2}(s+1)} = \frac{A}{s} + \frac{B}{s^{2}} + \frac{C}{s+1}
$$

Multiplying through by the denominator:
$$
s+3 = A(s)(s+1) + B(s+1) + C(s^{2})
$$

Setting $$ s = 0 $$:
$$
3 = B \implies B = 3
$$

Setting $$ s = -1 $$:
$$
-1 + 3 = C(1) \implies C = 2
$$

Setting $$ s = 1 $$:
$$
1 + 3 = A(1)(0) + 3(0) + 2(1) \implies 4 = 2 \implies A = 0
$$

Thus, we have:
\[
F(s) = \frac{3}{s^{2}} + \frac{2
The inverse Laplace transforms are: 1. $$ e^{-t} - e^{-2t} $$ 2. $$ \frac{1}{10} + \frac{1}{2}e^{-4t} + \frac{3}{5}e^{-5t} $$ 3. $$ e^{-t} - e^{-3t} $$ 4. $$ e^{-5t}\left(\cos(3t) + \frac{1}{3}\sin(3t)\right) $$ 5. $$ 3t + 2e^{-t} $$

Find the inverse Laplace transform of the following:

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