hi,

Inverse laplace

---(s + 9 )

------------

s^2 + 6s+ 9

-----(s + 3 )

--------------- + .........

(s + 3 )^ 2 + 4

{ e^-3t } * Inverse laplace { s / (s+4) } + .......

I have problem understand this step... how to take away the +3 in the equation ? i know there is a formula in the sheet to give a direct answer given but the example in the book really make my head crack (Headbang)

= e^-3t cos2t + ......

Thank you for any help :)