$I=2\int \dfrac{\sqrt{1-x}}{(1+x)^2. 2 \sqrt{1+x}}dx$ đặt $\sqrt{1+x}=t \Rightarrow \dfrac{1}{2\sqrt{1+x}}dx=dt$
$I=2\int \dfrac{\sqrt{2-t^2}}{t^4}dt$
Đặt $t=\sqrt 2 \cos u \Rightarrow dt=-\sqrt 2 \sin u du$
$I=-4\int \dfrac{ \sin^2 u}{4 \cos^4 u}du =-\int \tan^2 u d(\tan u)=...$