## Thomas' Calculus 13th Edition

$$\frac{\pi }{2}$$
\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}