$I = \int \dfrac{1}{\sqrt{x^2 -3x +1}}dx = \int \dfrac{1}{\sqrt{(x -\dfrac{3}{2})^2 -\dfrac{5}{4}}}dx $
đặt $x-\dfrac{3}{2} = \dfrac{5}{4\cos t} \Rightarrow dx = \dfrac{5\sin t}{4\cos^2 t}$
$I = \dfrac{5}{4}\int \dfrac{\sin t}{\cos^2 t . \sqrt{\dfrac{5}{4}\bigg (\dfrac{1}{\cos^2 t} - 1}\bigg )}dt$
$= \dfrac{5}{4}\int \dfrac{\sin t}{\cos^2 t . \sqrt{\dfrac{5}{4}\dfrac{1-\cos^2 t}{\cos^2 t}}}dt = \dfrac{5}{4}\int \dfrac{\sin t}{\cos^2 t . \sqrt{\dfrac{5}{4}\tan^2 t}}dt $
$=\dfrac{5\sqrt 5 }{10} \int \dfrac{\tan t}{\cos t \tan t}dt = \dfrac{5\sqrt 5 }{10} \int \dfrac{1}{\cos t}dt$
Tính $\int \dfrac{1}{\cos t}dt = \int \dfrac{\cos t dt}{\cos^2 t}=\int \dfrac{d(\sin t)}{1-\sin^2 t} = \int \dfrac{1}{1-u^2}du$
cái này dễ bạn tự làm nhé