jmc

Squaring the equation $x-\frac{1}{x}=3,$ we get $$x^2-2+\frac{1}{x^2}=9.$$Adding 4, we get $x^2+2+{\frac{1}{x}}^2=13,$ so $${\left(x+\frac{1}{x}\right)}^2=13.$$Since $x$ is positive, $$x+\frac{1}{x}=\boxed{\sqrt{13}}.$$