Finding the inverse laplace of this function: $ F(s)= \frac{s+8}{s^{2}+4s+5}$

Question

Im trying to find the inverse laplace of : $ F(s)= \frac{s+8}{s^{2}+4s+5}$

I reached the following: $$ F(s)= \frac{s}{(s+2)^{2}+1} + 8 \times \frac{1}{(s+2)^{2}+1}$$

Now i have the 2nd term in the right form, but my first one is not so im confused on how to go about putting it :S.

So basically in the first part how do i convert $ \frac{s}{(s+2)^{2}+1} to \frac{s+2}{(s+2)^{2}+1}$ ?

Answer 1

Hint. You may observe that $$\mathcal{L}(e^{-at}\cos\omega t) = \frac{s + a}{(s+a)^2 + \omega^2},$$ $$\mathcal{L}(e^{-at}\sin\omega t) = \frac{\omega}{(s+a)^2 + \omega^2}.$$ Then apply it to $$ F(s)=\frac{s+\color{red}{2}}{(s+\color{red}{2})^{2}+1^2}+6\:\times\frac{\color{blue}{1}}{(s+2)^{2}+\color{blue}{1}^2} . $$

Answer 2

Divide it as $s+2$ and $6$.

Hope that helps.