Factorize both sides of the equation:
[tex]\dfrac x{\sqrt{x^2+1}} = x^4 - x[/tex]
[tex]\dfrac x{\sqrt{x^2+1}} = x(x^3-1)[/tex]
[tex]\dfrac x{\sqrt{x^2+1}} - x(x^3-1) = 0[/tex]
[tex]x \left(\dfrac1{\sqrt{x^2+1}} - (x^3 - 1)\right) = 0[/tex]
One immediate solution is then x = 0. This leaves us with
[tex]\dfrac1{\sqrt{x^2+1}} - (x^3 - 1) = 0[/tex]
or as we had earlier,
[tex]\dfrac1{\sqrt{x^2+1}} = x^3 - 1[/tex]
Take the square of both sides:
[tex]\dfrac1{x^2+1} = \left(x^3-1\right)^2[/tex]
[tex]\dfrac1{x^2+1} = x^6 - 2x^3 + 1[/tex]
Turn this into a polynomial equation:
[tex]1 = \left(x^2+1\right) \left(x^6-2x^3+1\right)[/tex]
Expand the right side and make it equal to zero:
[tex]1 = x^8 + x^6 - 2x^5 - 2x^3 + x^2 + 1[/tex]
[tex]0 = x^8 + x^6 - 2x^5 - 2x^3 + x^2[/tex]
Each term in the polynomial has a common factor of x². Factoring this out just gives x = 0 again as a solution.
[tex]0 = x^2 \left(x^6 + x^4 - 2x^3 - 2x + 1\right)[/tex]
[tex]0 = x^6 + x^4 - 2x^3 - 2x + 1[/tex]
You'll need a computer to solve the remain se.xtic equation. Solving over reals, you would find two solutions, x ≈ 0.438 and x ≈ 1.18, but only x ≈ 1.18 is valid.
