## Thomas' Calculus 13th Edition

$$\pi$$
\eqalign{ & \int_2^4 {\frac{{2dx}}{{{x^2} - 6x + 10}}} \cr & {\text{complete the square for }}{x^2} - 6x + 10 \cr & = \left( {{x^2} - 6x + 9} \right) + 1 \cr & = {\left( {x - 3} \right)^2} + 1 \cr & = \int_2^4 {\frac{{2dx}}{{{{\left( {x - 3} \right)}^2} + 1}}} \cr & {\text{use the substitution method}}{\text{:}} \cr & u = x - 3,{\text{ so that }}du = dx \cr & {\text{the new limits on }}t{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}x = 4,{\text{ then }}u = 4 - 3 = 1 \cr & \,\,\,\,\,\,{\text{If }}x = 2,{\text{ then }}u = 2 - 3 = - 1 \cr & {\text{then}} \cr & \int_2^4 {\frac{{2dx}}{{{{\left( {x - 3} \right)}^2} + 1}}} = \int_{ - 1}^1 {\frac{{2du}}{{{u^2} + 1}}} \cr & = 2\int_{ - 1}^1 {\frac{{du}}{{{u^2} + 1}}} \cr & {\text{integrate by using the formula }}\int {\frac{{du}}{{{u^2} + {a^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\left( {\frac{u}{a}} \right) + C\,\,\,\left( {{\text{see page 419}}} \right)} \cr & {\text{with }}a = 1 \cr & = 2\left( {{{\tan }^{ - 1}}u} \right)_{ - 1}^1 \cr & = 2\left( {{{\tan }^{ - 1}}\left( 1 \right) - {{\tan }^{ - 1}}\left( { - 1} \right)} \right) \cr & = 2\left( {\frac{\pi }{4} + \frac{\pi }{4}} \right) \cr & = \pi \cr}