I used the residue theorem to solve the inverse Laplace transform of:
$$f(t)=\mathcal L^{-1}\Bigg( {s e^{zs} \over (k-s)^2(k+s)^2}\Bigg) \tag 1$$
where $k$ and $z$ are non-negative. I have two poles of order two at $k$ and $-k$. I calculated the first residual as:
$$R(k)={d \over ds}{se^{(z+t)s}\over(k+s)^2}\Bigg |_{s=k} = {e^{(t+z)s}\over(k+s)^3}\Big( (t+z)(ks+s^2)+k-s \Big) \Bigg |_{s=k}={(t+z)e^{k(t+z)} \over 4k}\tag 2$$
Similarly, I calculated the other residual as follows:
$$R(-k)={d \over ds}{se^{(z+t)s}\over(k-s)^2}\Bigg |_{s=-k} = {-e^{(t+z)s}\over(k-s)^3}\Big( (t+z)(s^2-ks)-k-s \Big) \Bigg |_{s=-k}=-{(t+z)e^{-k(t+z)} \over 4k}\tag 3$$
After adding them up I computed:
$$f(t) = R(k) + R(-k)={ (t+z)\big(e^{k(t+z)} - e^{-k(t+z)} \big) \over 4k} \tag 4$$
Wolfram seems to get my result multiplied with the Heaviside function:
$$f_{w}(t) =H(t+z){ (t+z)\big(e^{k(t+z)} - e^{-k(t+z)} \big) \over 4k} \tag 5$$
So my question is, where did the Heaviside function come from, and is my computation incorrect?