Question 5.

Let us consider 3 cases:

1) $x = 0$. This is a solution, as both L.H.S and R.H.S will be equal (0) when $x = 0$. (1 solution)

2) x > 0  

$\Rightarrow\ 2x\left(x^2+1\right)=5x^2$  

$\Rightarrow\ 2\left(x^2+1\right)=5x$  

$\Rightarrow\ 2x^2-5x+2=0\ \Rightarrow\ 2x^2-4x-x-2=0$  

$\Rightarrow\ 2x(x-2)-1(x-2)=0$  

$\Rightarrow\ (x-2)(2x-1)=0\ \Rightarrow\ x=2\ \text{or}\ \frac{1}{2}\ \Rightarrow$ (1 integer solution)

3) x < 0  

$\Rightarrow\ -2x\left(x^2+1\right)=5x^2$  

$\Rightarrow\ 2x^2+5x+2=0$  

$\Rightarrow\ 2x^2+4x+x+2=0$  

$\Rightarrow\ 2x(x+2)+1(x+2)=0$  

$\Rightarrow\ (x+2)(2x+1)=0\ \Rightarrow\ x=-2\ \text{or}\ -\frac{1}{2}\ \Rightarrow$ (1 integer solution)

So, the total number of integer solutions are $0,\,2,\,-2\ \Rightarrow\ 3$.