Laplace Inverse of $\frac{s+1}{s^2 + 2s}$

Question

Here I have $$F(s) = \frac{s+1}{s^2 + 2s}$$

Taking Laplace inverse on both sides,

$$\mathcal{L}^\prime \{F(s)\} = \mathcal{L}^\prime \left(\frac{s}{s^2+2s}\right) + \mathcal{L}^\prime \left(\frac{1}{s^2 +2s}\right)$$

$\therefore F(t) = e^{-2t} + ?$

I am unable to do it for extreme RHS terms, any way to do it.

Answer 1

You have to expand the given function in a partial fraction expansion

$ \dfrac{s+1}{s^2+2 s} = \dfrac{A}{s} + \dfrac{B}{s+2}$

It follows that $ s+1 = A (s+2) + B s $

whose solution is $ A = \frac{1}{2}, B = \frac{1}{2} $

Hence, the inverse Laplace transform of the right side is

$ F(t) = \frac{1}{2} (1 + e^{-2t} ) $

Answer 2

You can use $\mathcal{L}\{1-e^{-at}\} = \frac{\alpha}{s(s+\alpha)}$

Answer 3

$$F(s) = \frac{s+1}{s^2 + 2s}$$ $$F(s) = \frac{s+1}{(s+1)^2-1}$$ $$F(s-1) = \frac{s}{s^2-1}$$ $$F(s-1) = \mathcal{L}\left\{\cosh (t)\right\}$$ Apply inverse Laplace transform on both sides: $$e^{t} f(t) =\cosh (t)$$ $$\implies f(t) =\dfrac 12 (e^{-2t}+1)$$