Laplace Transform assistance Find the inverse laplace transform of: $\frac{25}{(s-1)^2(s^2+4)}$
$\frac{25}{(s-1)^2(s^2+4)}=\frac{A}{s-1}+\frac{B}{(s-1)^2}+\frac{C}{s^2 + 4}$
$$25=A(s^2+4)(s-1)+B(s^2+4)+C(s-1)^2$$
$\frac{25}{(s-1)^2(s^2+4)}=\frac{0}{s-1}+\frac{5}{(s-1)^2}+\frac{-15}{s^2 + 4}$
Maybe I don't know how to transform, but I got $5e^{\alpha t} -7.5 \sin(\alpha t)$ and this is apparently wrong. What is wrong?

$$25=A(s^2+4)(s-1)+B(s^2+4)+(Cs+D)(s-1)^2$$ is correct. Therefore we have $$25=s^3(A+C)+s^2(-A+B-2C+D)+s(4A+C-2D)+(-4A+4B+D)$$ Now, $$A=0, -A+B+C=0,4A-2C=0, -4A+4B+C=25$$ which gives $$A=-2, C=2, D=-3, B=5$$
