jmc

We look for integers $a$ and $b$ such that $$\sqrt{41+24\sqrt{2}}=a+b\sqrt{2}.$$Squaring both sides, we have $41+24\sqrt{2}=(a+b\sqrt{2}{)}^2=(a^2+2b^2)+2ab\sqrt{2}.$ Therefore, we must have $$\begin{array}{rl}{a}^2+2b^2& =41,\\ 2ab& =24.\end{array}$$The second equation gives $ab=12.$ Trying the factor pairs of $12,$ we find that $(a,b)=(3,4)$ satisfies $a^2+2b^2=41.$ Therefore, $41+24\sqrt{2}=(3+4\sqrt{2}{)}^2.$ Since $3+4\sqrt{2}\ge 0,$ it follows that $$\sqrt{41+24\sqrt{2}}=\boxed{3+4\sqrt{2}}.$$