Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 8 - Integration Techniques, L'Hopital's Rule, and Improper Integrals - 8.2 Exercises: 11


$-\frac 1 {16e^{4x}}(1+4x)+C$

Work Step by Step

$dv=e^{-4x}dx$ $v=\int e^{-4x}dx = -\frac 1 4e^{-4x}$ $u=x, du=dx$ $\int xe^{-4x}dx = x(-\frac 1 4 e^{-4x})-\int -\frac 1 4e^{-4x}dx$ $\frac x 4 e^{-4x}- \frac 1 {16}e^{-4x}+C$ $-\frac 1 {16e^{4x}}(1+4x)+C$
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.