I want to find the function $x(t)$ which unilatteral Laplace transform is:
$$\mathcal X(p) = \frac{a}{p(p+b)} \tag 1$$
I know that: $$\mathcal L_u^{-1}\left(\frac{1}{p+a}\right) = e^{-at}\nu(t)\tag 2$$
And:
$$\mathcal L_u\left(\frac{dx}{dt}\right) = p\mathcal X(p) \tag 3$$
So my idea is to "reverse" $(3)$ and to suppose the "$1/p$" means an integral:
$$x(t) = a\int_{-\infty}^t e^{-bu}\nu(t) \ du \tag 4$$
Is this correct and thorough or is their a problem ?
Let be $$ X(p)=\frac{a}{p(p+b)}=\frac{F(p)}{p}\Longrightarrow x(t)=\int_0^tf(\tau)\,\mathrm d\tau $$ and $$ f(t)=\mathcal L^{-1}\left(\frac{a}{p+b}\right)=a\,\mathrm e^{-bt}u(t) $$ where $u(t)$ is the Heaviside step function. So we have $$ x(t)=\int_0^tf(\tau)\,\mathrm d\tau=a\int_0^t\mathrm e^{-b\tau}u(\tau)\,\mathrm d\tau=a\int_0^t\mathrm e^{-b\tau}\,\mathrm d\tau=\frac{a}{b}\left(1- e^{-bt}\right)u(t) $$
The same result with partial fraction décomposition $$ X(p)=\frac{a}{p(p+b)}=\frac{a}{b}\left(\frac{1}{p}-\frac{1}{p+b}\right)\Longrightarrow x(t)=\frac{a}{b}\left(1- e^{-bt}\right)u(t) $$