Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.2 Integration by Parts - 7.2 Exercises - Page 520: 15

Answer

\[ = \frac{{{x^3}\ln x}}{3} - \frac{1}{9}{x^3} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {{x^2}\ln x\,dx} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ u = \ln x\,\,\,\,\,\, \to \,\,\,\,\,du = \frac{{dx}}{x} \hfill \\ dv = {x^2}dx\,\,\, \to \,\,\,v = \frac{{{x^3}}}{3} \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = \frac{{{x^3}\ln 3}}{3} - \int_{}^{} {\,\left( {\frac{{{x^3}}}{3}} \right)\,\left( {\frac{{dx}}{x}} \right)} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = \frac{{{x^3}\ln x}}{3} - \frac{1}{3}\int_{}^{} {{x^2}dx} \hfill \\ \hfill \\ integrate\,\,{\text{using }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \hfill \\ \hfill \\ = \frac{{{x^3}\ln x}}{3} - \frac{1}{9}{x^3} + C \hfill \\ \end{gathered} \]
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