jmc

We can write $$\frac{x^2+7}{\sqrt{x^2+3}}=\frac{x^2+3+4}{\sqrt{x^2+3}}=\frac{x^2+3}{\sqrt{x^2+3}}+\frac{4}{\sqrt{x^2+3}}=\sqrt{x^2+3}+\frac{4}{\sqrt{x^2+3}}.$$By AM-GM, $$\sqrt{x^2+3}+\frac{4}{\sqrt{x^2+3}}\ge 2\sqrt{\sqrt{x^2+3}\cdot \frac{4}{\sqrt{x^2+3}}}=4.$$Equality occurs when $x=1,$ so the minimum value is $\boxed{4}.$