jmc

When we expand this out, we get a bunch of terms with $\sqrt{7}$ in them. To avoid painful estimation, we do the following trick: Add $(5-2\sqrt{7}{)}^4$ to this expression. We know that $(5-2\sqrt{7}{)}^4$ is small, since $2\sqrt{7}=\sqrt{28}$ is close to $5=\sqrt{25}$, at least compared to $6=\sqrt{36}$. When we add these together, the $\sqrt{7}$ terms magically cancel out. By the Binomial Theorem, $$(5+2\sqrt{7}{)}^4=5^4+4\cdot 5^3\cdot (2\sqrt{7})+6\cdot 5^2\cdot (2\sqrt{7}{)}^2+4\cdot 5\cdot (2\sqrt{7}{)}^3+(2\sqrt{7}{)}^4$$ whereas $$(5-2\sqrt{7}{)}^4=5^4-4\cdot 5^3\cdot (2\sqrt{7})+6\cdot 5^2\cdot (2\sqrt{7}{)}^2-4\cdot 5\cdot (2\sqrt{7}{)}^3+(2\sqrt{7}{)}^4.$$ Therefore, their sum is $$2(5^4+6\cdot 5^2(2\sqrt{7}{)}^2+(2\sqrt{7}{)}^4)=2(625+4200+784)=11218.$$ Since the term we added, $(5-2\sqrt{7}{)}^4$, is less than a half (actually, it's less than .01), $\boxed{11218}$ is the closest integer to $(5+2\sqrt{7}{)}^4$.