Find inverse laplace transform for x(s) = (2)/(s(s+1)^2)

Answer:x(t)=[tex]2-2e^{-t}-2te^{-t}[/tex]Step-by-step explanation:The function x(s) = (2)/(s(s+1)^2) can be expressed as partial fractions:X(s)=[tex]\frac{2}{s(s+1)^2}=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{(s+1)^2}[/tex]2=A·(s+1)²+B·s·(s+1)+C·s2=A·(s²+2s+1)+B·(s²+s)+C·s2=A·s²+2sA+A+B·s²+Bs+C·s2=s²(A+B)+s(2A+B+C)+ASo we can find the values of A, B and C by solving these equations:A+B=0 ⇒ 2+B=0 ⇒ B= -22A+B+C=0 ⇒ 2·2+(-2)+C=0 ⇒ C= -2A=2So X(s) is expressed as:X(s)=[tex]\frac{2}{s}+\frac{-2}{s+1}+\frac{-2}{(s+1)^2}[/tex]Using the inverse laplace transform tables we obtain x(t):ℒ⁻¹{X(s)}=ℒ⁻¹{[tex]\frac{2}{s}+\frac{-2}{s+1}+\frac{-2}{(s+1)^2}[/tex]}ℒ⁻¹{X(s)}=ℒ⁻¹{[tex]\frac{2}{s}[/tex]}+ℒ⁻¹{[tex]\frac{-2}{s+1}[/tex]}+ℒ⁻¹{[tex]\frac{-2}{(s+1)^2}[/tex]}x(t)=[tex]2-2e^{-t}-2te^{-t}[/tex]