$I=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}}\frac{1+\sin2x+\cos2x}{\sin x+\cos x}dx$ $=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}}\frac{(\sin x +\cos x)^2+(\sin x +\cos x)(\cos x - \sin x)}{\sin x+\cos x}dx$
$=\int\limits_{\frac{\pi }{6}}^{\frac{\pi }{2}}2\cos xdx$
$=2\sin x \bigg|_ \frac{\pi }{6}^\frac{\pi }{2}=1.$