## Inverse Laplace Transform - Step Function?

Given...

(2s$$^{2}$$+1)e$$^{-s}$$
------------------
(s - 1)(s$$^{2}$$ + 4s + 5)

Find the inverse Laplace Transform.

I am unsure where to start and am just looking for a little direction, not an answer. From past experience, I am assuming I will have to rewrite this in another form...perhaps by taking out the e$$^{-s}$$ term. I also believe I will need to complete the square on the quadratic in the denominator.

If someone could just point me in the right direction, that would be greatly appreciated! Thanks!!!
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 Recognitions: Science Advisor First, use the properties of the Laplace transform to get rid of the $e^{-s}$. Then use partial fractions to do the rest.
 Ok, so I did that...here's my work...(note, I factored out the 1/10 that appears in both terms). (e$$^{-s}$$/10)$$\frac{17s+5}{(s+2)^{2}+4s+5}$$+$$\frac{3}{s-1}$$ Then I completed the square on the quadratic in the denominator and rewrote it... (e$$^{-s}$$/10)$$\frac{17s+5}{(s+2)^{2}+1}$$+$$\frac{3}{s-1}$$ Now, finding the inverse Laplace Transform of 3/(s-1) is very simple. So is finding the inverse transform of e$$^{-s}$$. But the term containing the 17s + 5 is confusing me.

Recognitions:

## Inverse Laplace Transform - Step Function?

Try writing

$$\frac{17s}{(s+2)^2+1} + \frac{5}{(s+2)^2+1}$$

Also, your formulas will be easier to read if you write the entire thing in TeX instead of just bits and pieces.
 Recognitions: Science Advisor And you should be able to get rid of the $e^{-s}$ part by using this property: $$\mathcal{L}\{f(t-a)\} = e^{-as}F(s)$$
 Thanks very much! This is only my second time using TeX so I apologize for that. Getting used to it.