x^2+sqrt(x^2+11)<31

$sqrt{x^{2}+11}=t, t geq 0$

$x^{2}=t^{2}-11$

$t^{2}-11+t-31<0$

$t^{2}+t-42<0$

$(t+7)(t-6)<0$

$-7 < t < 6$

$-7 < sqrt{x^{2}+11} < 6$

$0 leq sqrt{x^{2}+11} < 6$

$left { {{sqrt{x^{2}+11} geq 0} atop {sqrt{x^{2}+11} < 6}} right.$

$left { {{x^{2}+11 geq 0} atop {x^{2}+11 < 36}} right.$

$x^{2}-25<0$

$(x-5)(x+5)<0$

$-5 <x< 5$

Ответ: $x in (-5;5)$