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

 A: 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 $x=1,-1$
A: $\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}$
Away from $x\in \{-1,0,1\}$ it suffices to find the solutions of $f(x)=0$, where
$$f(x)\stackrel{\text{def}}{=}\abs{x^2+x}-\abs{x^2-1}-\abs{x-1}\text{.}$$
As stated in the comments, there are four cases to work through:
$$f(x)=2\cdot
\begin{cases}
x & x < -1 \\
-1 & -1 < x < 0 \\
x^2 + x -1 & 0 < x < 1 \\
1 & 1 < x
\end{cases}\text{.}$$
But this case analysis shows that $f$ is nondecreasing and has exactly one root, in the interval $(0,1)$. Therefore the only integral solution of the original equation is the trivial one at $x=0$.
A: Easier solution and a more analytical solution using the property of modulus:
$$|x| + |\frac{x}{x - 1}|=\frac{x^2}{|x - 1|}$$
$$\text{As, } |a| + |b| = |a + b| \text{ only, when } ab \geq 0$$
$$\text{and } |x| + |\frac{x}{x - 1}| = |x + \frac{x}{x - 1}|$$ is true only, if $x\cdot \frac{x}{x-1} \geq 0$
$$\implies x\in {0} \cup (1, \infty)$$
$\therefore$ Least value of x = 0
A: Since $0$ is a solution, the least integer solution has to be $\le0$, so $|x|=-x$ and $|x-1|=1-x$.
Thus the equation becomes
$$
-x-\frac{x}{1-x}=\frac{x^2}{1-x}
$$
that is
$$
-x+x^2-x=x^2
$$
Hence $x=0$ is the only solution $\le0$.
A: Case $3$ is not as many cases as you'd think.
Consider.
Case 1: $x \ge 1$ then $x - 1 \ge 0$ and and the modulus are non-negitive.  However $\frac x{x-1}$ is undefined for $x = 1$ so $x> 1$
$x + \frac x{x-1} = \frac {x(x-1) + x}{x-1} = \frac {x^2}{x-1}$ which is always true so $x > 1$ are solutions.
Case 2: $x \le 0$ so $x -1 < 0$ but $\frac {x}{x-1} \ge  0$.
So $-x + \frac x{x-1} = - \frac {x^2}{x-1}$ which is always true if 
$\frac {-x(x-1) + x}{x-1} = \frac {-x^2 + 2x}{x-1} = \frac {-x^2}{x-1}$ which only happens if so $x = 0$.  So $x = 0$ is   solution.
Case 3: $0 < x < 1$ then $x > 0$ and $x-1 < 0$ and $\frac x{x-1} < 0$.
So we want $x - \frac x{x-1} = -\frac {x^2}{x-1}$ which is true if
$\frac {x(x-1) - x}{x-1} = \frac {-x^2}{x-1}$ which is true if
$\frac {x^2 - 2x}{x-1} = \frac {-x^2}{x-1}$ which is true if and only if
$x^2 - 2x = -x^2$
$2x^2 = 2x$ 
$x^2 = x$ which happens if $x = 0$ or $x =1$ but those are out of range.
So solutions are $x= 0$ or $x > 1$.
A: Alternatively:
$$|x| + |\frac{x}{x - 1}| = \frac{x^2}{|x - 1|} \iff \\
|x|\left(1+\frac{1}{|x-1|}\right)=\frac{|x|^2}{|x-1|} \iff \\
1) \ |x|=0 \ \ \text{or} \ \ 2) \ 1+\frac1{|x-1|}=\frac{|x|}{|x-1|} \iff \\
1) \ x=0;\\
2) \ \frac{|x-1|+1}{|x-1|}=\frac{|x|}{|x-1|} \stackrel{x\ne 1}{\Rightarrow} \\
|x-1|=|x|-1 \Rightarrow x> 1.$$
Hence, the solution of the original equation is $x\in \{0\}\cup (1,+\infty)$ and the least integer solution is $x=0$.
