## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 7 - Integration Techniques - 7.4 Trigonometric Substitutions - 7.4 Exercises: 61

#### Answer

$$\frac{{x - 1}}{2}\sqrt {{x^2} - 2x + 10} - \frac{9}{2}\ln \left| {\frac{{x - 1 + \sqrt {{x^2} - 2x + 10} }}{3}} \right| + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{{x^2} - 2x + 1}}{{\sqrt {{x^2} - 2x + 10} }}} dx \cr & = \int {\frac{{{{\left( {x - 1} \right)}^2}}}{{\sqrt {{x^2} - 2x + 10} }}} dx \cr & {\text{completing the square}} \cr & = \int {\frac{{{{\left( {x - 1} \right)}^2}}}{{\sqrt {{x^2} - 2x + 1 + 9} }}} dx = \int {\frac{{{{\left( {x - 1} \right)}^2}}}{{\sqrt {{{\left( {x - 1} \right)}^2} + 9} }}} dx \cr & u = x - 1,{\text{ }}du = dx \cr & = \int {\frac{{{{\left( {x - 1} \right)}^2}}}{{\sqrt {{{\left( {x - 1} \right)}^2} + 9} }}} dx = \int {\frac{{{u^2}du}}{{\sqrt {{u^2} + 9} }}} \cr & {\text{substitute }}u = 3\tan \theta ,{\text{ }}du = 3{\sec ^2}\theta d\theta \cr & = \int {\frac{{{u^2}du}}{{\sqrt {{u^2} + 9} }}} = \int {\frac{{\left( {9{{\tan }^2}\theta } \right)\left( {3{{\sec }^2}\theta d\theta } \right)}}{{\sqrt {9{{\tan }^2}{\theta ^2} + 9} }}} = \int {\frac{{\left( {9{{\tan }^2}\theta } \right)\left( {3{{\sec }^2}\theta d\theta } \right)}}{{3\sqrt {{{\tan }^2}\theta + 1} }}} \cr & {\text{pythagorean identity}} \cr & = \int {\frac{{\left( {9{{\tan }^2}\theta } \right)\left( {3{{\sec }^2}\theta d\theta } \right)}}{{3\sqrt {{{\sec }^2}\theta } }}} = 9\int {{{\tan }^2}\theta \sec \theta } d\theta = 9\int {\left( {{{\sec }^2}\theta - 1} \right)\sec \theta } d\theta \cr & 9\int {\left( {{{\sec }^3}\theta - \sec \theta } \right)} d\theta \cr & {\text{integrating}} \cr & = 9\int {\left( {{{\sec }^3}\theta - \sec \theta } \right)d\theta } \cr & = 9\left( {\frac{1}{2}\sec \theta \tan \theta + \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right| - \ln \left| {\sec \theta + \tan \theta } \right|} \right) + C \cr & = 9\left( {\frac{1}{2}\sec \theta \tan \theta - \frac{1}{2}\ln \left| {\sec \theta + \tan \theta } \right|} \right) + C \cr & = \frac{9}{2}\left( {\sec \theta \tan \theta - \ln \left| {\sec \theta + \tan \theta } \right|} \right) + C \cr & u = 3\tan \theta \cr & = \frac{9}{2}\left( {\left( {\frac{{\sqrt {{u^2} + 9} }}{3}} \right)\left( {\frac{u}{3}} \right) - \ln \left| {\frac{{\sqrt {{u^2} + 9} }}{3} + \frac{u}{3}} \right|} \right) + C \cr & = \frac{u}{2}\sqrt {{u^2} + 9} - \frac{9}{2}\ln \left| {\frac{{u + \sqrt {{u^2} + 9} }}{3}} \right| + C \cr & u = x - 1 \cr & = \frac{{x - 1}}{2}\sqrt {{{\left( {x - 1} \right)}^2} + 9} - \frac{9}{2}\ln \left| {\frac{{x - 1 + \sqrt {{{\left( {x - 1} \right)}^2} + 9} }}{3}} \right| + C \cr & = \frac{{x - 1}}{2}\sqrt {{x^2} - 2x + 10} - \frac{9}{2}\ln \left| {\frac{{x - 1 + \sqrt {{x^2} - 2x + 10} }}{3}} \right| + C \cr}

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