Answer
$$x = \pm \root 4 \of 2 $$
Work Step by Step
$$\eqalign{
& \frac{{2\left( {{x^2} - 1} \right)\sqrt {{x^2} + 1} - \frac{{{x^4}}}{{\sqrt {{x^2} + 1} }}}}{{{x^2} + 1}} = 0 \cr
& {\text{Multiply the numerator and denominator by }}\sqrt {{x^2} + 1} \cr
& \frac{{2\left( {{x^2} - 1} \right){{\left( {\sqrt {{x^2} + 1} } \right)}^2} - \frac{{{x^4}\sqrt {{x^2} + 1} }}{{\sqrt {{x^2} + 1} }}}}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }} = 0 \cr
& {\text{Simplify}} \cr
& \frac{{2\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right) - {x^4}}}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }} = 0 \cr
& \frac{{2\left( {{x^4} - 1} \right) - {x^4}}}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }} = 0 \cr
& \frac{{2{x^4} - 2 - {x^4}}}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }} = 0 \cr
& \frac{{{x^4} - 2}}{{\left( {{x^2} + 1} \right)\sqrt {{x^2} + 1} }} = 0 \cr
& {\text{Cross product}} \cr
& {x^4} - 2 = 0 \cr
& x = \pm \root 4 \of 2 \cr} $$
