$I=\int_1^2 \dfrac{\sqrt{x^2-1}.x}{x^2}dx$ đặt $\sqrt{x^2-1}=t \Rightarrow x dx =tdt$
$I=\int_0^{\sqrt 3} \dfrac{t.t}{t^2+1}dt =\int \bigg (1-\dfrac{1}{t^2+1}\bigg)dt=t \bigg |_0^{\sqrt 3} -\int \dfrac{1}{t^2+1}dt=\sqrt 3 -I_1$
Tính $I_1$ Đặt $t=\tan u \Rightarrow dt=\dfrac{1}{\cos^2 u}du=(1+\tan^2 u)du=(1+t^2)du$
$\Rightarrow du =\dfrac{dt}{1+t^2} \Rightarrow I_1=\int \dfrac{1}{t^2+1}dt =\int_0^{\frac{\pi}{3}} du=u \bigg |_0^{\frac{\pi}{3}}=\dfrac{\pi}{3}$
Vậy $I=\sqrt 3 -\dfrac{\pi}{3}$