Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.1 Integration by Parts - Exercises - Page 395: 7


$$-(4x-3)e^{-x}- 4 e^{-x} +c .$$

Work Step by Step

We do the integration by parts; we choose $u=4x-3$ and $dv=e^{-x}dx$. Then, $du=4dx$, $v=-e^{-x}$ $$\int (4x-3)e^{-x}dx=\int udv=uv-\int vdu\\ =-(4x-3)e^{-x}+ 4\int e^{-x} dx\\ =-(4x-3)e^{-x}- 4 e^{-x} +c .$$
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