straightforward question: How do I find the inverse laplace of (1/(s^2+1)^2)?

Question

anonymous · Answer

If it helps, the answer is 1/2*sin(t)-1/2*t*cos(t), but how do I get that answer?

anonymous · Answer

Use F(s)=2bs/(s^2+b^2)^2. Everythign fits that form except for the 2, so you multiple the top and bottom by 2. Factor out the bottom 2 to get the 1/2, and then plug in the f(t) from table

anonymous · Answer

? There is no s in the numerator

anonymous · Answer

oops sorry i copied wrong one down from table, i meant 2b^3/(s^2+b^2)^2

anonymous · Answer

1. Which table are you looking at? Mine only contains very basic and common transforms. and 2. Is there a way to do it using more simple table references? (e.g. using only knowledge of basic Laplace properties and knowledge of Laplace transforms of t, sint, and cost, using partial fractions or something similar?)

anonymous · Answer

http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx it's number 11 on this table, and it showed up on mine from class as well i dont know how to do it any other way, it has no partial fractions sorry