In my book I have to do the inverse Laplace transform of: $$\displaystyle \frac{4 + 15 s}{s(2+5s)} = \frac{15}{5} \frac{\left(s + \frac{4}{15}\right)}{s\left(s + \frac{2}{5}\right)}$$
With a partial fraction division, I can find that one fraction is $\displaystyle\frac{2}{s}$ and the second $\displaystyle\frac{5}{5s+2}$ and I can find $e^{-\frac{2}{5}t} + 2$.
But the book states that I should get it directly by looking at the summary of usual Laplace transforms? I checked the formulas and did not find what I should be able to see. Can someone help me to navigate the classical formula table given that equation?
Here. Formula $14.$
Write $\frac{15 s+4}{s (5 s+2)}$ as $$\frac{15}{5} \frac{\left(s + \frac{4}{15}\right)}{s\left(s + \frac{2}{5}\right)}$$ $\alpha=\frac{4}{15};\;a=0;\;b=\frac{2}{5}$ $$3\cdot \frac{1}{\frac{2}{5}} \left[\frac{4}{15}e^0-\left(\frac{4}{15}-\frac{2}{5}\right)e^{-2/5t}\right]=\frac{15}{2}\left(\frac{4}{15}+\frac{2}{15}e^{-2/5t}\right)=2+e^{-2/5t}$$
Tbh I prefer your solution :)