## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 8 - Further Techniques and Applications of Integration - 8.1 Integration by Parts - 8.1 Exercises - Page 432: 14

#### Answer

$$\frac{1}{6}\ln \left| {2{x^3} + 1} \right| + C$$

#### Work Step by Step

\eqalign{ & \int {\frac{{{x^2}dx}}{{2{x^3} + 1}}} \cr & \cr & {\text{integrate by substitution}} \cr & {\text{set }}u = 2{x^3} + 1{\text{ then }}\frac{{du}}{{dx}} = 6{x^2} \to {x^2}dx = \frac{1}{6}du \cr & {\text{write the integrand in terms of }}u \cr & \int {\frac{{{x^2}dx}}{{2{x^3} + 1}}} = \int {\frac{{\left( {1/6} \right)du}}{u}} \cr & = \frac{1}{6}\int {\frac{{du}}{u}} \cr & {\text{integrating}} \cr & = \frac{1}{6}\ln \left| u \right| + C \cr & {\text{replace }}2{x^3} + 1{\text{ for }}u \cr & = \frac{1}{6}\ln \left| {2{x^3} + 1} \right| + C \cr}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.