My Slightly Overcomplicated Problem

Given expression can be written as

$\dfrac{(x+1)(x^{2}+1-x)}{(x+1)(x-1)}=\dfrac{x^{2}+1-x}{x-1}=x+\dfrac{1}{x-1}=x+\sqrt{\dfrac{6}{x}}$

So $\dfrac{1}{x-1}=\sqrt{\dfrac{6}{x}}$

$\Rightarrow x-1=\sqrt{\dfrac{x}{6}}$

Whole square and re-arrange to get $x^{2}-\dfrac{13}{6}x+1=0$

$\large=(x-\dfrac{3}{2})( x-\dfrac{2}{3})$

But $x =\dfrac{2}{3}$ does not satisfy the equation.

So the answer should be $\dfrac{3}{2}$