Chapter 7 - Review - Exercises - Page 553: 27

$\sqrt 2+3\ln (\sqrt 2+1)$

$$\eqalign{ & \int_0^2 {\sqrt {{x^2} - 2x + 2} } dx \cr & {\text{Completing the square}} \cr & = \int_0^2 {\sqrt {\left( {{x^2} - 2x + 1} \right) + 1} } dx \cr & = \int_0^2 {\sqrt {{{\left( {x - 1} \right)}^2} + 1} } dx \cr & {\text{From the triangle shown below}} \cr & \tan \theta = x - 1 \to dx = {\sec ^2}\theta + 1 \cr & \theta = {\tan ^{ - 1}}\left( {x - 1} \right) \cr & \sqrt {{{\left( {x - 1} \right)}^2} + 1} = \sec \theta \cr & {\text{Therefore}}{\text{,}} \cr & \int {\sqrt {{{\left( {x - 1} \right)}^2} + 1} } dx = \int {\sec \theta } \left( {{{\sec }^2}\theta + 1} \right)d\theta \cr & = \int {\left( {{{\sec }^3}\theta + \sec \theta } \right)} d\theta \cr & {\text{Integrating}}{\text{, use the entry 71 and 14}}{\text{, table of integrals}} \cr & {\text{ = }}\frac{1}{2}\sec \theta \tan \theta + \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right| + \ln \left| {\sec \theta + \tan \theta } \right| + C \cr & {\text{ = }}\frac{1}{2}\sec \theta \tan \theta + \frac{3}{2}\ln \left| {\sec \theta + \tan \theta } \right| + C \cr & {\text{Write in terms of }}x{\text{ using the triangle}} \cr & {\text{ = }}\frac{1}{2}\left( {x - 1} \right)\sqrt {{{\left( {x - 1} \right)}^2} + 1} + \frac{3}{2}\ln \left| {\sqrt {{{\left( {x - 1} \right)}^2} + 1} + x - 1} \right| + C \cr & {\text{ = }}\frac{1}{2}\left( {x - 1} \right)\sqrt {{x^2} - 2x + 2} + \frac{3}{2}\ln \left| {\sqrt {{x^2} - 2x + 2} + x - 1} \right| + C \cr & {\text{Therefore}}{\text{,}} \cr & \int_0^2 {\sqrt {{x^2} - 2x + 2} } dx \cr & = \left[ {\frac{1}{2}\left( {x - 1} \right)\sqrt {{x^2} - 2x + 2} + \frac{3}{2}\ln \left| {\sqrt {{x^2} - 2x + 2} + x - 1} \right|} \right]_0^2 \cr & {\text{Evaluating the limits}} \cr & = \left[ {\frac{1}{2}\left( {2 - 1} \right)\sqrt {{2^2} - 2\left( 2 \right) + 2} + \frac{3}{2}\ln \left| {\sqrt {{2^2} - 2\left( 2 \right) + 2} + 2 - 1} \right|} \right] \cr & = - \left[ {\frac{1}{2}\left( {0 - 1} \right)\sqrt 2 + \frac{3}{2}\ln \left| {\sqrt 2 + 0 - 1} \right|} \right] \cr & = \left( {\frac{{\sqrt 2 }}{2} + \frac{3}{2}\ln \left( {1 + \sqrt 2 } \right)} \right) - \left( { - \frac{{\sqrt 2 }}{2} + \frac{3}{2}\ln \left( { - 1 + \sqrt 2 } \right)} \right) \cr & = \frac{{\sqrt 2 }}{2} + \frac{3}{2}\ln \left( {1 + \sqrt 2 } \right) + \frac{{\sqrt 2 }}{2} - \frac{3}{2}\ln \left( { - 1 + \sqrt 2 } \right) \cr & = \sqrt 2 + \frac{3}{2}\ln \left( {\frac{{\sqrt 2 + 1}}{{\sqrt 2 - 1}}} \right) \cr & = \sqrt 2 + \frac{3}{2}\ln \left( {\frac{{(\sqrt 2 + 1)(\sqrt 2-1}}{{(\sqrt 2 +1)(\sqrt 2-1)}}} \right) \cr & = \sqrt 2 + \frac{3}{2}\ln \left( {\frac{{(\sqrt 2 + 1)(\sqrt 2+1}}{{(\sqrt 2 -1)(\sqrt 2+1)}}} \right) \cr & = \sqrt 2 + \frac{3}{2}\ln \left( \sqrt 2 + 1 \right)^2 \cr &=\sqrt 2+3\ln (\sqrt 2+1)} $$