## Calculus: Early Transcendentals (2nd Edition)

Published by Pearson

# Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises: 30

#### Answer

$= 8 + 3\ln 3$

#### Work Step by Step

$\begin{gathered} \int_2^4 {\frac{{{x^2} + 2}}{{x - 1}}dx} \hfill \\ \hfill \\ use\,\,the\,\,long\,\,\,division \hfill \\ \hfill \\ = \int_2^4 {\,\left( {x + \frac{{x + 2}}{{x - 1}}} \right)dx = \int_2^4 {\,\left( {x + \frac{{x - 1 + 3}}{{x - 1}}} \right)dx} } \hfill \\ \hfill \\ = \int_2^4 {\,\left( {x + 1 + \frac{3}{{x - 1}}} \right)dx} \hfill \\ \hfill \\ integrate\,and\,use\,\,the\,\,ftc \hfill \\ \hfill \\ = \,\,\left[ {\frac{{{x^2}}}{2} + x + 3\ln \,\left( {x - 1} \right)} \right]_2^4 \hfill \\ \hfill \\ = 8 + 4 + 3\ln 3 - \,\left( {2 + 2 + 3\ln 1} \right) \hfill \\ \hfill \\ {\text{Simplify}} \hfill \\ \hfill \\ = 8 + 3\ln 3 \hfill \\ \end{gathered}$

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