Answer
$$\frac{\pi }{2}$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^{ - 1} {\frac{{2dv}}{{{v^2} + 4v + 5}}} \cr
& = \int_{ - 2}^{ - 1} {\frac{2}{{{v^2} + 4v + 4 + 1}}dv} \cr
& = \int_{ - 2}^{ - 1} {\frac{2}{{{{\left( {v + 2} \right)}^2} + 1}}dv} \cr
& {\text{Use the substitution method:}} \cr
& u = v + 2,\,\,\,\,\,\,du = dv \cr
& {\text{The new limits on }}u{\text{ are found as follows}} \cr
& \,\,\,\,\,\,{\text{If }}v = - 1,{\text{ then }}u = - 1 + 2 = 1 \cr
& \,\,\,\,\,\,{\text{If }}v = - 2,{\text{ then }}u = - 2 + 2 = 0 \cr
& {\text{Then}} \cr
& \int_{ - 2}^{ - 1} {\frac{2}{{{{\left( {v + 2} \right)}^2} + 1}}dv} = 2\int_0^1 {\frac{{du}}{{{u^2} + 1}}} \cr
& {\text{integrate}} \cr
& = 2\left( {{{\tan }^{ - 1}}u} \right)_0^1 \cr
& {\text{use the fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 281}}} \right) \cr
& = 2\left( {{{\tan }^{ - 1}}1 - {{\tan }^{ - 1}}0} \right) \cr
& {\text{simplifying, we get:}} \cr
& = 2\left( {\frac{\pi }{4}} \right) \cr
& = \frac{\pi }{2} \cr} $$
