How to solve x'+x=delta(x) with Laplace transform.

Question

I'm solving the following ODE: $$x'+x=\delta(t)\tag{1}$$ where $\delta(x)$ is the Dirac delta function. By Laplace transform, I get $$sX(s)-x(0)+X(s)=1.$$ Hence, $$X(s)=\frac{x(0)+1}{s+1}\tag{2}$$ Now implement the inverse transform: $$x(t)=[x(0)+1]\exp(-t)\tag{3}$$ However, this is obviously not correct because when I let $t=0$ in formula (3), I get $x(0)=x(0)+1$ I want to know where I made a mistake, and how to solve it correctly by Laplace Transform. Thanks in advance!