Differential equation laplace transform

Find the inverse laplace transform of

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

( 1 / ( s * (s^2+2s+2) ) )

( 1 / ( s * (s^2+2s+2) ) ) = [ A / s ] + [ B *s + C / ( s^2+2s+2) ]

1 = A * ( s^2+2s+2) + (B * s + C) * s

1 = A * s^2 + 2 * A * s + 2 * A + B * s^2 + C * s

1 = 2 * A ===> A = 1/2

[ 0 = A + B ] * s^2 ====> (-1/2) = B

[ 0 = 2 * A + C ] * s ====> C = -2 * A ===> C = -2 * (1/2) ===> C = -1


( 1 / ( s * (s^2+2s+2) ) ) = [ (1/2) / s ] + [ ((-1/2) * s - 1) / ( s^2+2s+2) ]


( 1 / ( s * (s^2+2s+2) ) ) = (1/2) * (1 / s) + [ ((-1/2) * s ) / ( s^2+2s+2) ] - [ 1 / ( s^2+2s+2) ]

( 1 / ( s * (s^2+2s+2) ) ) = (1/2) * (1 / s) - (1/2) * [ s / ( s^2+2s+2) ] - [ 1 / ( s^2+2s+2) ]


let's do each one separately;

ℒ⁻¹{(1/2) * (1 / s) } = (1/2) * 1 = (1/2)


ℒ⁻¹{ - (1/2) * [ s / ( s^2+2s+1 + 1) ] }
ℒ⁻¹{ - (1/2) * [ s / (s + 1)^2 + 1) ] }
ℒ⁻¹{ - (1/2) * [ (s + 1)-1 / (s + 1)^2 + 1) ] }
ℒ⁻¹{ - (1/2) * ( [ (s + 1) / (s + 1)^2 + 1) ] - ( 1 / (s + 1)^2 + 1) ) }
ℒ⁻¹{ - (1/2) * [ (s + 1) / (s + 1)^2 + 1) ] + (1/2) * ( 1 / (s + 1)^2 + 1) ) }
- (1/2) * e^(-t) * cos( t ) + (1/2) * e^(-t) * sin( t )


ℒ⁻¹{ - [ 1 / ( (s +1)^2 + 1 ) ] } = - e^(-t) * sin(t)


the answer:

(1/2) - (1/2) * e^(-t) * cos( t ) + (1/2) * e^(-t) * sin( t ) - e^(-t) * sin(t)

(1/2) - (1/2) * e^(-t) * cos( t ) - (1/2) * e^(-t) * sin( t )