Can the Inverse Finding the Laplace transform of $\frac{2s + 1}{s(s + 1)(s + 2)}$ without using partial fractions?

Question

I'm wondering if we can perhaps using the convolution theorem to find the inverse Laplace transform of $\dfrac{2s + 1}{s(s + 1)(s + 2)}$? I can find it using partial fraction decomposition, but it is not obvious to me whether this is the convolution of two functions?

Thank you for any help.

Answer 1

It is not right away the convolution of two functions but you can split into two fractions and use convolution on each one and add the results .

Answer 2

$$F(s)=\dfrac{2s + 1}{s(s + 1)(s + 2)}$$ With $s\to s-\dfrac12$ $$\dfrac{2s}{\left(s^2-\dfrac14\right)\left(s+\dfrac32\right)}=\dfrac{s}{\left(s^2-\dfrac14\right)}\dfrac{2}{s+\dfrac32}=\int_0^x\cosh\dfrac12t\cdot 2e^{-\frac32t}\ dt=\int_0^x e^{-t}+e^{-2t}\ dt=-e^{-x}-\dfrac12e^{-2x}+\dfrac32$$ then from $F(s-c)=e^{cx}f(x)$ we have $$f(x)=e^{-\frac12x}\left(-e^{-x}-\dfrac12e^{-2x}+\dfrac32\right)=-e^{-\frac32x}-\dfrac12e^{-\frac52x}+\dfrac32e^{-\frac12x}$$