$\int\limits_{0}^{1}\frac{dx}{\sqrt{x+1}+\sqrt{3(1+x)}}$.Đặt $u=\sqrt{x+1} \Rightarrow du=\frac{1}{2\sqrt{x+1}}dx$
Đổi cận: $x=0 \Rightarrow u=1; x=1 \Rightarrow u=\sqrt{2}$
$I=\int\limits_{1}^{\sqrt{2}}\frac{2u}{u+\sqrt{3}u}du=2\int\limits_{1}^{\sqrt{2}}\frac{u}{u+\sqrt{3}u}du=2\int\limits_{1}^{\sqrt{2}}\frac{1}{1+\sqrt{3}}du=2\left[ {\frac{1}{1+\sqrt{3}}x} \right]_{1}^{\sqrt{2}} =(\sqrt{6}-\sqrt{2})-(-1+\sqrt{3})=\sqrt{6}-\sqrt{2}-\sqrt{3}+1$