Find the inverse laplace transform of complicated rational function 
Find the inverse laplace $$\mathcal{L}^{-1}\left\{ \frac{2s + 1}{2s^2 + s + 2} \right\}$$

I don't see a simple way to take the inverse laplace transform, convolution theorem? 

This is a case where you "massage" the expression to fit it into a more standard form
First get rid of the leading coefficient in the denominator.
$$ \frac{2s+1}{2s^2+s+2}=\frac{s+\frac{1}{2}}{s^2+\frac{1}{2}s+1}$$
Then complete the square on the denominator.
$$ \frac{s+\frac{1}{2}}{s^2+\frac{1}{2}s+1}=\frac{s+\frac{1}{2}}{\left(s+\frac{1}{4}\right)^2+\frac{15}{16}} $$
Then re-write the numerator to obtain the result
$$ \frac{2s+1}{2s^2+s+2}= \frac{\left(s+\frac{1}{4}\right)+\frac{1}{4}}{\left(s+\frac{1}{4}\right)^2+\frac{15}{16}} $$
Set this up for solving in terms of $\sin$ and $\cos$ and $e^{-t/4}$
$$ \mathcal{L}^{-1}\left(\frac{\left(s+\frac{1}{4}\right)}{\left(s+\frac{1}{4}\right)^2+\left(\frac{\sqrt{15}}{4}\right)^2}\right)+\frac{1}{4}\cdot\frac{4}{\sqrt{15}}\mathcal{L}^{-1}\left(\frac{\frac{\sqrt{15}}{4}}{\left(s+\frac{1}{4}\right)^2+\left(\frac{\sqrt{15}}{4}\right)^2}\right) $$
to obtain the result
$$f(t)=\left[\cos\left(\frac{\sqrt{15}}{4}t\right)+\frac{\sqrt{15}}{15}\sin\left(\frac{\sqrt{15}}{4}t\right)\right]e^{-t/4}$$
