câu 3 nhé
Đặt $x=-t \Rightarrow dx=-dt$
$I=-\int_1^{-1}\dfrac{1}{1-t+\sqrt{t^2+1}}dt = \int_{-1}^1 \dfrac{1}{1-x+\sqrt{x^2+1}}dx$
$\Rightarrow 2I = \int_{-1}^1 \bigg ( \dfrac{1}{(1+x)+\sqrt{x^2+1}} + \dfrac{1}{(1-x)+\sqrt{x^2+1}} \bigg )dx =\int_{-1}^1 dx =x \bigg |_{-1}^1 =2$
$\Rightarrow I = 1$