3, \pi
$\int\limits_{\dfrac{2\sqrt{3} -3}{2}}^{\dfrac{-1}{2}}\dfrac{1}{(2x+3)^{2}+4}dx$
Đặt (2x+3)=2tan t$\Rightarrow dx=(tan^{2}x+1)dt$
x=$\frac{2\sqrt{3}-3 }{2}\Rightarrow t=\frac{\pi }{3}$
x=-0,5$\Rightarrow t=\frac{\pi }{4}$
$\int\limits_{\pi /3}^{\pi /4}\frac{tan^{2}+1}{4tan^{2}x+4}dt=\int\limits_{\pi /3}^{\pi /4}\frac{1}{4}dt=\frac{-\pi }{48}$