The least integer value of $x$, which satisfy $|x| + |\frac{x}{x - 1}| = \frac{x^2}{|x - 1|}$, is... 
Question:  The least integer value of $x$, which satisfy $|x| + |\frac{x}{x - 1}| = \frac{x^2}{|x - 1|}$, is...


MY ATTEMPT:
Case 1: Taking all the modulus positive and solving ...
Case 2: Taking all the modulus negative and solving...
Case 3: Taking different signs of different modulus at a time and solving...
But the Case 3 will arise to many different cases and this will take a lot of time to solve. So, definitely an alternate method is required. Please provide an efficient method to solve this question. Thank you!
P.S.: The answer to this question is x=0

This is a new edit made 8 hours ago. Anyone who posted their answers before that are not wrong and not flawed. Their answer will satisfy this question: The least integer value of $x$, which satisfy $|x| + |\frac{x}{x + 1}| = \frac{x^2}{|x - 1|}$, is.... Sorry for the confusion


This is probably not the most efficient way of doing it, but in my opinion it is more fun
$$|x|+\frac{|x|}{|x+1|}=\frac{x^2}{|x-1|}$$
$$x^2+\frac{x^2}{(x+1)^2}+\frac{2x^2}{|x+1|}=\frac{x^4}{(x-1)^2}$$
So clearly $x=0$ is a solution. Taking out a factor of $x^2$, we get
$$1+\frac{1}{(x+1)^2}+\frac{2}{|x+1|}=\frac{x^2}{(x-1)^2}$$
$$\frac{x^2}{(x-1)^2}-\frac{1}{(x+1)^2}-1=\frac{2}{|x+1|}$$
$$\frac{x^4}{(x-1)^4}+\frac{1}{(x+1)^4}+1-\frac{x^2}{(x-1)^2(x+1)^2}-\frac{x^2}{(x-1)^2}+\frac{1}{(x+1)^2}=\frac{4}{(x+1)^2}$$
$$\frac{3}{(x+1)^2}=\frac{x^4(x+1)^4+(x-1)^4+(x+1)^4(x-1)^4-x^2(x+1)^2(x-1)^2-x^2(x-1)^2(x+1)^4}{(x-1)^4(x+1)^4}$$
So if we say $x\neq-1,1$, we can multiply through by $(x+1)^4(x-1)^4$:
$$3(x-1)^4(x+1)^2=x^4(x+1)^4+(x-1)^4+(x+1)^4(x-1)^4-x^2(x+1)^2(x-1)^2-x^2(x-1)^2(x+1)^4$$
So it will be the solutions of this polynomial except for $x=1,-1$
