## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 7 - Integration - 7.2 Substitution - 7.2 Exercises - Page 375: 20

#### Answer

$\frac{1}{3}\ln \left| {{t^3} + 6t + 3} \right| + C$

#### Work Step by Step

$\begin{gathered} \int_{}^{} {\frac{{{t^2} + 2}}{{{t^3} + 6t + 3}}dt} \hfill \\ Let\,\,u = {t^3} + 6t + 3\,\,,\,So\,\,that \hfill \\ du = \,\left( {3{t^2} + 6} \right)dt \hfill \\ du = 3\,\left( {{t^2} + 2} \right)dt \hfill \\ \int_{}^{} {\frac{{{t^2} + 2}}{{{t^3} + 6t + 3}}dt\,\, = \frac{1}{3}\int_{}^{} {\frac{{3\,\left( {{t^2} + 2} \right)dt}}{{{t^3} + 6t + 3}}} } \hfill \\ \frac{1}{3}\int_{}^{} {\frac{{du}}{u}} \hfill \\ Integrating \hfill \\ \frac{1}{3}\ln \left| u \right| + C \hfill \\ Substituting\,\,u = {t^3} + 6t + 3\,\,gives \hfill \\ \frac{1}{3}\ln \left| {{t^3} + 6t + 3} \right| + C \hfill \\ \end{gathered}$

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