I have tried to solve that equation $x^{x^3-x}=2^{x^2+x}$ in $\mathbb{R}$ , I have got only two integers solutions which they are : $x=-1$, $x=2$ , are there others ?
Note: if we try to study this : $\frac{x^3-x}{x^2+x}=\frac{\log x}{\log 2}$, I think there is a numerical solution in the range $( 0,1)$ using value intermediate theorem, it is to show that there is a solution here, but i can't determine it
On
I think the domain should be restricted to positive real numbers. I mean, the expressions make sense for some negative numbers, but not so many of them. (It is problematic to raise a negative number to a non-integer exponent.)
So let $x$ be positive. Then after taking the logarithm we obtain: $(x^2+x)\log 2= \log x\cdot (x^3-x)$. Simplyfying by $x$ and $x+1$ yields $\log 2= \log x\cdot (x-1)$, or equivalently $\log 2- \log x\cdot (x-1) =0$. Clearly $x=2$ is a solution. The function $f(x)= \log 2- \log x\cdot (x-1)$ on the left is strictly monotone increasing on $]0,1[$ and decreasing on $]1,\infty[$ (pay attention to the signs). As $f(1)>0$, there can be two roots, one on each interval. There is one on $]1,\infty[$, namely $x=2$. There is also one on $]0,1[$, as $f$ changes signs there. For example $f(1/4)< 0$.
HINT: Write your equation in the form $$\ln(x)-\frac{x^2+x}{x^3-1}\ln(2)=0$$, define
$$f(x)=\ln(x)-\frac{x^2+x}{x^3-1}\ln(2)$$ and use calculus.