$I=\int \dfrac{\sqrt{(2x+3)^2 -4}}{2x+3} dx$ đặt $2x+3 =\dfrac{2}{\cos t}$
$\Rightarrow dx = \sin t(1+\tan^2 t)dt$
$I=\int \dfrac{\sqrt{4(1+\tan^2 t) -4}}{\dfrac{2}{\cos t}} .\sin t (1+\tan^2 t) dt$
$=\int_0^{\frac{\pi}{3}} \tan t \sin t \cos t (1+\tan^2 t) dt =\int \dfrac{\sin^2 t }{\cos^2 t}dt=\int \dfrac{1}{\cos^2 t}dt -\int dt$
$=\tan t -t + C$