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Đặt $t = \sqrt {{e^x} - 1} \Rightarrow {t^2} = {e^x} - 1 \Rightarrow
2tdt = {e^x}dx \Rightarrow dx = \frac{{2tdt}}{{{e^x}}} =
\frac{{2tdt}}{{{t^2} + 1}} $
$\Rightarrow I = \int\limits_0^1
{\frac{{2{t^2}}}{{{t^2} + 1}}} dt = 2\int\limits_0^1 {\left( {1 -
\frac{1}{{{t^2} + 1}}} \right)} dt =2\left ( t - \arctan x \right )
|_0^1= \boxed{\frac{{4 - \pi }}{2}} $
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