I know $$\mathcal{L}\{f'(t)\}=\frac{7s^3-41s^2+84s}{(s-2)(s^2-4s+13)}$$ and I have to find $f(t)$, given $f(0)=4$.
Since $$\mathcal{L}\{f'(t)\}=s\mathcal{L}\{f(t)\}-f(0)$$ then $$f(t)=\mathcal{L}^{-1}\left\{\frac{1}{s}\cdot\frac{7s^3-41s^2+84s}{(s-2)(s^2-4s+13)}-\frac{4}{s}\right\}$$ i.e., $$f(t)=\mathcal{L}^{-1}\left\{\frac{7s^2-41s+84}{(s-2)(s^2-4s+13)}-\frac{4}{s}\right\}$$
By partial fractions, I got $$\frac{7s^2-41s+84}{(s-2)(s^2-4s+13)}=\frac{10}{3}\cdot\frac{1}{s-2}+\frac{11}{3}\cdot\frac{s-2}{(s-2)^2+3^2}-\frac{13}{3}\cdot\frac{3}{(s-2)^2+3^2}$$ Therefore, $$f(t)=\frac{10}{3}e^{2t}+\frac{11}{3}e^{2t}\cos(3t)-\frac{13}{3}e^{2t}\sin(3t)+4$$ because $$\mathcal{L}^{-1}\left\{\frac{4}{s}\right\}=4$$
But,it's clear $f(0)\neq4$ and also, when I found $f'(t)$ using the expression above, I didn't get the same answer to $\mathcal{L}\{f'(t)\}$. What is wrong?
The issue is that $f'$ is not regular. Looking at the original Laplace transform, is $f'(t) = 7\delta(t)+g(t)$ with $g$ a regular function. You can see this by inspecting the degree of the rational function which is $1$ for $f'$. Thus, even if you start at $f(0)=4$, due to the Dirac delta, $f(0^+)=4+7=11$, which is consistent with your final result.
Hope this helps.