Đặt $x=-t$ ta có
$I=\int_{-1}^{1}\frac{1}{1-t+\sqrt{t^{2}+1}}dt=\int_{-1}^{1}\frac{1}{1-x+\sqrt{x^{2}+1}}dx$
$\Rightarrow 2I = \int_{-1}^{1}\frac{1}{1+x+\sqrt{x^{2}+1}}dx+\int_{-1}^{1}\frac{1}{1-x+\sqrt{x^{2}+1}}dx$
$= \int_{-1}^{1} \dfrac{1+x+\sqrt{x^{2}+1}+1-x+\sqrt{x^{2}+1}}{(1+\sqrt{x^2+1})^2 -x^2}dx =\int_{-1}^1 dx = 2 \Rightarrow I=1$