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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?

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    $\begingroup$ When you see a quadratic polynomial like $2s^2+s+2$ that has a first-degree term, you should often think of completing the square. $\endgroup$ – Michael Hardy Jul 30 '17 at 2:08
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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

$$f(t)=\left[\cos\left(\frac{\sqrt{15}}{4}t\right)+\frac{\sqrt{15}}{15}\sin\left(\frac{\sqrt{15}}{4}t\right)\right]e^{-t/4}$$

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