Find the solution of $1 \le \left\lfloor \left\lvert \frac{x+1}{x-3}\right\rvert +x\right\rfloor \lt 2$

Find the solution of $$1 \le \left\lfloor \left\lvert \frac{x+1}{x-3}\right\rvert +x\right\rfloor \lt 2$$ My try:

The only thing i know is that $$\left\lfloor \left\lvert \frac{x+1}{x-3}\right\rvert +x\right\rfloor \in \Bbb Z$$ so $$\left\lfloor \left\lvert \frac{x+1}{x-3}\right\rvert +x\right\rfloor = 1$$

but i don't know how to continue.

Any hints?

Hint: $$1 \leq \lfloor x \rfloor <2$$ iff $$1 \leq x <2$$ so you have to solve $$1-x \leq |\frac {1+x} {x-3}| < 2-x$$. Consider the cases $$\frac {1+x} {x-3}>0$$ and $$\frac {1+x} {x-3} \leq 0$$ separately.