Number of integers satisfying the equation $|x^2+5x|+|x-x^2|=\left|6x\right|\text{\hspace{0.17em}is }\mathrm{.......}$

 $\left|x\right|+\left|y\right|=|x+y|\Rightarrow x,y\text{\hspace{0.17em}should have same sign.}$
Case (i): $x\ge 0,y\ge 0$ 
$\therefore x^2+5x\ge 0\text{\hspace{0.17em}and\hspace{0.17em}}x-x^2\ge 0$ 
$x(x+5)\ge 0\text{\hspace{0.17em}and\hspace{0.17em}}x(1-x)\ge 0$ 
$\Rightarrow x(x+5)\ge 0\text{\hspace{0.17em}and\hspace{0.17em}}x(x-1)\le 0$ 
$x\in (-\infty ,-5]\cup [0,\infty )\&x\in [0,1]$ 
$\Rightarrow x=0,1$ 
Case: (ii): $x\le 0\&y\le 0$ 
$\Rightarrow x^2+5x\le 0\text{\hspace{0.17em}and\hspace{0.17em}}x-x^2\le 0$ 
$\Rightarrow x\in [-5,0]\text{\hspace{0.17em}and\hspace{0.17em}}x\in (-\infty ,0]\cup [1,\infty ]$ 
$\Rightarrow x=-5,-4,-3,-2,-1,0,1$ 
So total 7 distinct integral values of x.