Tính $I=\int\limits_{1}^{2} \frac{1}{x^2\sqrt{x^2+1}}dx$

Tính $I=\displaystyle\int\limits_{1}^{2} \dfrac{1}{x^2\sqrt{x^2+1}}dx$

Đặt $t=\dfrac{1}{x} \Rightarrow -dt=\dfrac{1}{x^2}dx$
$x=1 \Rightarrow t=1 \\ x=2 \Rightarrow t=\dfrac{1}{2}$

$I= \displaystyle\int\limits_{1}^{\frac{1}{2}} \dfrac{-1}{\sqrt{\dfrac{1}{t^2}+1}}dt  = \displaystyle\int\limits_{\frac{1}{2}}^{1} \dfrac{t}{\sqrt{t^2+1}}dt$

Đặt $u= \sqrt{t^2+1}$