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: 16

Answer

\[ = \frac{{{x^2}\ln x}}{2} - \frac{{{x^2}}}{4} + C\]

Work Step by Step

\[\begin{gathered} \int_{}^{} {x\ln x\,\,dx} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ u = \ln x\,\,\,\,\,\,\, \to \,\,\,\,\,\,du = \frac{{dx}}{x} \hfill \\ dv = xdx\,\,\, \to \,\,\,\,\,\,\,\,v = \frac{{{x^2}}}{2} \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = \frac{{{x^2}}}{2}\ln x - \int_{}^{} {\,\left( {\frac{{{x^2}}}{2}} \right)\,\left( {\frac{{dx}}{x}} \right)} \hfill \\ \hfill \\ integrate\,\,{\text{using }}\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} \hfill \\ \hfill \\ = \frac{{{x^2}\ln x}}{2} - \frac{{{x^2}}}{4} + C \hfill \\ \end{gathered} \]
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.