I can't find the inverse Laplace of : $$\frac{p_1p_2}{(x+p_1)(x+p_2)}$$
I remove first $p_1,p_2$ and try to $$\mathscr{L}^{-1}\frac{1}{(x+p_1)(x+p_2)}$$ $$\frac{1}{(x+p_1)(x+p_2)}= \frac{A}{(x+p_1)} + \frac{B}{(x+p_2)}$$ $$1 = A(x+p_2) + B(x+p_1)$$
With $x = 0$
$$1 = Ap_2 + Bp_1$$
Then with $x = 1$ and $x = -1$, I get: $$1 = A + B \\ A = 1-B \\ 1 = (1-B)p_2 + Bp_1\\ 1 = p_2 -Bp_2 + Bp_1 \\ B = \frac{1-p_2}{p_1-p_2} \\ A = 1- \frac{1-p_2}{p_1-p_2} = \frac{p_1-1}{p_1-p_2}$$
Which would mean that $$\frac{p_1-1}{p_1-p_2} * \frac{p_1p_2}{x+p_1} + \frac{1-p_2}{p_1-p_2}\frac{p_1p_2}{x+p_2}$$ So I see that we should have $e^{-p_1x}$ and $e^{-p_2x}$ at some point but I don't know what to do with the factors $\frac{p_1-1}{p_1-p_2}$ and $\frac{1-p_2}{p_1-p_2}$. What is wrong and why?
Edit: I am adding a picture of the reply.
$$\frac{p_1 p_2}{(p_1+x) (p_2+x)}=\frac{A}{p_1+x}+\frac{B}{p_2+x}=\frac{Ap_2+Ax+Bp_1+Bx}{(p_1+x) (p_2+x)}$$
Therefore it must be $$\frac{(Ap_2+Bp_1)+(A+B)x}{(p_1+x) (p_2+x)}=\frac{(p_1 p_2)+0\cdot x}{(p_1+x) (p_2+x)}$$ this means $$ \begin{cases} Ap_2+Bp_1=p_1p_2\\ A+B=0\\ \end{cases} $$ $$ \begin{cases} Ap_2-Ap_1=p_1p_2\to A(p_2-p_1)=p_1p_2\\ B=-A\\ \end{cases} $$ $$ \begin{cases} A=\frac{p_1p_2}{p_2-p_1}\\ B=-\frac{p_1p_2}{p_2-p_1}\\ \end{cases} $$ therefore $$\frac{p_1 p_2}{(p_1+x) (p_2+x)}=\frac{p_1p_2}{p_2-p_1}\left(\frac{1}{p_1+x}-\frac{1}{p_2+x}\right)$$ Inverse Laplace transform is $$\frac{p_1p_2}{p_2-p_1} \left(e^{-p_1 t}-e^{-p_2 t}\right)$$