Solve the following equation: $\sqrt{x^2+x}-\sqrt{x^2-x}=x+1$ I have tried many ways but not success
$\sqrt{x^2+x}-\sqrt{x^2-x}=x+1$
$\Leftrightarrow \sqrt{x}(\sqrt{x+1}-\sqrt{x-1})=x+1$
or $\sqrt{x^2+x}-\sqrt{x^2-x}=x+1 \Leftrightarrow \sqrt{\dfrac{x}{x+1}}-\dfrac{\sqrt{x(x-1)}}{x+1}=1 \Leftrightarrow \dfrac{x}{x+1}=1+\dfrac{2\sqrt{x(x-1)}}{x+1}+\dfrac{x(x-1)}{(x+1)^2} \Leftrightarrow \dfrac{(x+1)^2+2\sqrt{x(x-1)}(x+1)+x(x-1)-x(x+1)}{(x+1)^2}=0 \Leftrightarrow x^2+1+2\sqrt{x(x-1)}(x+1)=0 $
I need your help!

Using your first approach, square both sides:
$$x \left((x+1) - 2\sqrt{x^2-1} + (x-1)\right) = (x+1)^2.$$  Now rearrange and collect like terms to isolate the square root:
$$2x \sqrt{x^2 - 1} = 2x^2 - (x+1)^2 = x^2 - 2x - 1.$$
Next, square again:
$$4x^2 (x^2 - 1) = (x^2 - 2x - 1)^2.$$
The resulting quartic has a rather complicated closed form, but it does exist; the only thing left is to eliminate extraneous roots.
