[tex]\displaystyle \\ \\ \int\limits_{2}^4 \dfrac{3}{3x^2+12x+24} \, dx = \int\limits_{2}^4 \dfrac{3}{3(x^2+4x+8)} \, dx = \int\limits_{2}^4 \dfrac{1}{x^2+4x+8} \, dx = \\ \\ =\int\limits_{2}^4 \dfrac{1}{x^2+4x+4+4} \, dx = \int\limits_{2}^4 \dfrac{1}{(x+2)^2+2^2} \, dx = \\ \\ = \int\limits_{2}^4 \dfrac{(x+2)'}{(x+2)^2+2^2} \, dx = \left| \dfrac{1}{2}\cdot \text{arctg}\Big( \dfrac{x+2}{2} \Big)\right|_2^4= \\ \\ = \dfrac{1}{2}\cdot \Big[\text{arctg}\Big( \dfrac{4+2}{2} \Big) - \text{arctg}\Big( \dfrac{2+2}{2} \Big)\Big] =[/tex]
[tex]=\dfrac{\text{arctg}( 3 ) - \text{arctg}( 2 )}{2}[/tex]
