## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises: 33

#### Answer

$= \frac{1}{3}{\tan ^{ - 1}}\,\left( {\frac{{x - 1}}{3}} \right) + C$

#### Work Step by Step

$\begin{gathered} \int_{}^{} {\frac{{dx}}{{{x^2} - 2x + 10}}} \hfill \\ \hfill \\ complete\,\,the\,\,square\,\,and\,\,factor \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{{x^2} - 2x + 1 + 9}}} \hfill \\ \hfill \\ = \int_{}^{} {\frac{{dx}}{{\,{{\left( {x - 1} \right)}^2} + 9}}} \hfill \\ \hfill \\ set\,\,\left( {x - 1} \right) = u\,\, \to \,\,dx = du \hfill \\ \hfill \\ {\text{Therefore}}{\text{,}} \hfill \\ \hfill \\ \int_{}^{} {\frac{{dx}}{{{x^2} - 2x + 10}}} = \int_{}^{} {\frac{{du}}{{{u^2} + 9}}} \hfill \\ \hfill \\ Usi\,ng\,\int_{}^{} {\frac{{dx}}{{{x^2} + {a^2}}} = \frac{1}{a}{{\tan }^{ - 1}}\frac{x}{a}} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \frac{1}{3}{\tan ^{ - 1}}\frac{u}{3} + c \hfill \\ \hfill \\ substituting\,\,back\,\,u = \,\left( {x - 1} \right) \hfill \\ \hfill \\ = \frac{1}{3}{\tan ^{ - 1}}\,\left( {\frac{{x - 1}}{3}} \right) + C \hfill \\ \end{gathered}$

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