Example of Inverse Laplace's Transform for repeated poles using residual method 

Okkkkkk

It works as long as there aren't quadratic terms, or other irreducible polynomials beyond linear terms. Otherwise, you get interdependent unknown coefficients.

can you solve this task?

​@@SaninSelimovic-zh8eq Given: F(s) = (s^2+2*s+1)/(s*(s^2+2*s+5)) Complete the square on the denominator: F(s) = (s^2+2*s+1)/(s*((s+1)^2+4)) Set up partial fractions for a linear denominator term (A/s), and a quadratic denominator term (linear numerator), using the shifted value of s instead of just s. You'll see why this helps: F(s) = A/s + (B*(s + 1) + C)/((s+1)^2 + 4) Heaviside coverup finds A for us, at s=0: A = (0^2+2*0+1)/(covered*(0^2+2*0+5)) = 1/5 Reconstruct what remains: (s^2+2*s+1)/(s*((s+1)^2+4)) = 1/5/s + (B*(s + 1) + C)/((s+1)^2+4) Let s = -1, to solve for C. Notice that B cancels out: ((-1)^2+2*(-1)+1)/((-1)*((-1+1)^2+4)) = 1/5/(-1) + (B*((-1) + 1) + C)/((-1+1)^2+4) 0 = -1/5 + C/4 C = 4/5 Reconstruct what remains: (s^2+2*s+1)/(s*((s+1)^2+1)) = 1/5/s + (B*(s + 1) + 4/5)/((s+1)^2+4) Let s = -2, to solve for B. We choose -2, so all the (s+1)^2 terms become 1. ((-2)^2+2*(-2)+1)/(-2*((-2+1)^2+4)) = 1/5/(-2) + (B*(-2 + 1) + 4/5)/((-2+1)^2+4) -1/10 = -1/10 + (-B+ 4/5)/2 Solve for B: B = 4/5 Partial fraction result: F(s) = 1/5*[1/s + (4*(s + 1) + 4)/((s+1)^2+4)] Arrange to look like standard Laplace transforms: F(s) = 1/5*[1/s + 4*(s + 1)/((s+1)^2+4) + 2*2/((s+1)^2+4)] Inverse Laplace: f(t) = 1/5 + 1/5*e^(-t)*[4*cos(2*t) + 2*sin(2*t)]