Solving $x|x-1|<3$? Seems pretty straight-forward but I'm a bit confused with the modulus and inequality.
So I know that if

$|x-1|>3$

because the entire equation on the left is within the modulus we can write

$x-1>3$ or $x-1<-3$ 

I also know that if $y<0$

$y|x-1|>3$

Can be written as

$|x-1|<3/y$

But what do we do with 

$x|x-1|<3$

Can we break it down further without trying to input values?
 A: Let's do it like this. For $x>1$ the equation is equivalent to $x(x-1)<3$ this gives $x^2-x-3<0$ Roots of this equation are $$\frac{1+\sqrt{13}}{2}, \frac{1-\sqrt{13}}{2}$$
So this gives the range $x\in(1, \frac{1+\sqrt{13}}{2})$
For $x<1$ the equation is $-x(x-1)<3$ or $x^2-x+3>0$ this always satisfied for any values of $x$. Hence $x\in(-\infty, \frac{1+\sqrt{13}}{2})$, taking $x$ to be integers would make this $x\le2$.
A: This method doesn't generalize, but it may be instructive. The inequality surely holds for $x\le0$, so we can concentrate on $x>0$. Dividing both sides by $x$, we obtain
$$
|x-1|<\frac{3}{x}
$$
that's equivalent to
$$
-\frac{3}{x}<x-1<\frac{3}{x}
$$
that in turn becomes
\begin{cases}
x^2-x+3 > 0 \\[4px]
x^2-x-3 < 0
\end{cases}
The top one is true for every $x$; the bottom one is satisfied for
$$
0<x<\frac{1+\sqrt{13}}{2}
$$
(recall we're restricting to $x>0$).
Putting back the interval $x\le0$ we discussed before, we can conclude the solution set is
$$
x<\frac{1+\sqrt{13}}{2}
$$
A: Consider the cases where $x > 1$ and where $x < 1$.
If $x > 1$, we have:
$$x(x-1) < 3$$
$$\Rightarrow x^2-x-3 < 0$$
With $x=2$, we have $2^2-2-3 = -1 < 0$, and with $x = 3$ we have $3^2 - 3 - 3 = 3 > 0$, so $x ≤ 2$.
If $x < 1$, we have:
$$x(x+1) < 3$$
$$\Rightarrow x^2+x-3 < 0$$
Can you continue?
A: Consider the graph of $y=x(x-1)$:
$\hspace{2cm}$
If the absolute values are applied to get $y=x|x-1|$, the graph becomes:
$\hspace{2cm}$
The point $A$ has coordinates $x(x-1)=3 \Rightarrow x=\frac{1+\sqrt{13}}{2}$ and $y=3$. 
Hence, the solution of the inequality $x|x-1|<3$ is:
$$x\in \left(-\infty,\frac{1+\sqrt{13}}{2}\right).$$
