Question: $$\frac{x}{\sqrt{x^2+1}} = x^4 - x$$
I tried: $$\rightarrow \frac{1}{\sqrt{x^2+1}} = x^ 3 - 1$$
$$\to\frac{\sqrt{x^2 + 1}+1}{\sqrt{x^2+1}} = x^3$$
Now rationalising it $$\to \frac{x^2 +1-1}{x^2+1-\sqrt{x^2+1}} = x^3$$
$$\frac{1}{x^2+1-\sqrt{x^2+1}} = x$$
Can we do anything with this? How to solve? Please help me.
From the graph of the function : $$f(x) = \dfrac{1}{x^2 + 1 - \sqrt{x^2 + 1}} - x$$ we see that there is only one other solution $x = 1.181$.
We deduce that solutions of the equation are $0$ and $1.181$.