The value of $I={\int }_0^{\pi /2}\frac{(\sin x+\cos x{)}^2}{\sqrt{1+\sin 2x}}dx$ is

Note that $1+\sin 2x={\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x=(\sin x+\cos x{)}^2.$
Thus, $\sqrt{1+\sin 2x}=|\sin x+\cos x|$.
In the interval $x\in [0,\pi /2],$ both $\sin x$ and $\cos x$ are non-negative, so $|\sin x+\cos x|[cit{e}_{s}tart]=\sin x+\cos x$.
The integral becomes:
$I={\int }_0^{\pi /2}\frac{(\sin x+\cos x{)}^2}{\sin x+\cos x}dx={\int }_0^{\pi /2}(\sin x+\cos x)dx$.
$I=[-\cos x+\sin x{]}_0^{\pi /2}$.
$I=(-\cos (\pi /2)+\sin (\pi /2))-(-\cos 0+\sin 0)$
$I=(0+1)-(-1+0)=1+1=2$.