Đặt $x=\sin t \Rightarrow dx =\cos t dt$
$I=\int_0^{\frac{\pi}{2}} \dfrac{\cos t}{1+\sqrt{1-\sin^2 t}}dt =\int \dfrac{\cos t}{1+\cos t}dt =\int dt -\int \dfrac{1}{1+\cos t}dt$
$=t -\dfrac{1}{2}\int \dfrac{1}{\cos^2 \dfrac{t}{2}}dt=t-\int \dfrac{1}{\cos^2 \dfrac{t}{2}} d(\dfrac{t}{2})= (t-\tan \dfrac{t}{2} ) \bigg |_0^{\frac{\pi}{2}}=...$