What is the Inverse Laplace Transform of (s+1)/ (s^2 + 4s + 5) + e^-2s / 3s^4?

[math_04]

Homework Statement

Find the inverse laplace transform of (s+1)/ (s^2 + 4s + 5) + e^-2s / 3s^4

Homework Equations

The Attempt at a Solution

For the first one

(s+1) / (s^2 + 4s + 5), I completed the square for the denominator so

(s+1) / [(s+1)^2 + 1]

Now it gets confusing, how do I find out the inverse laplace of this.

And for e^-2s / 3s^4 , is the inverse laplace 1/9 x 1/ (s+2)^4 = 1/9(s+2)^4

 

[HallsofIvy]

The same way you find the inverse Laplace transform of just about anything: look it up in a table.
If you don't have one in your book, here's one:
http://www.vibrationdata.com/Laplace.htm

You will notice that they give an inverse Laplace transform for $s/(s^2+ \alpha$ as well as the inverse Laplace transform for $F(x-\alpha)$. Use them together for $(s+1)/((s+1)^2+ 2)$.

 

[math_04]

The problem is I don't get the LaTeX Code: s/(s^2+ \\alpha thing. I only get the f(x +a) thing.

 

[dirk_mec1]

The problem is I don't get the LaTeX Code: s/(s^2+ \\alpha thing. I only get the f(x +a) thing.

In the last line of Ivy you should recognize a form that can be found in the table. Ivy rewrote the equation in that form by taking out a square.