[tex]\displaystyle \int_{-2}^{-1}\dfrac{1}{x^2+4x+5}\,dx = \int_{-2}^{-1}\dfrac{1}{x^2+4x+4+1}\, dx =\\ \\ = \int_{-2}^{-1}\dfrac{1}{(x+2)^2+1}\, dx = \int_{-2}^{-1}\dfrac{(x+2)'}{(x+2)^2+1^2}\, dx =\\ \\ = \dfrac{1}{1}\arctan\dfrac{x+2}{1}\Big|_{-2}^{-1} = \arctan(1)-\arctan(0)=\\ \\ =\dfrac{\pi}{4} - 0 = \boxed{\dfrac{\pi}{4}}[/tex]
