University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.1 - Integration by Parts - Exercises - Page 427: 7



Work Step by Step

We use the formula: $\int f(x)g(x)dx=$ $f(x)\int g(x)dx-\int [f'(x)\int g(x)dx]dx$. Put f(x)=x and g(x)=$ e^{x}$. The derivative of first function is 1 and the integral of the second function is $e^{x}$. Therefore, $\int xe^{x}dx= xe^{x}-\int 1·e^{x}dx= xe^{x}-e^{x}+C$
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.