Calculus (3rd Edition)

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

Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 51


$$ \frac{-(x+1)}{3}e^{4-3 x} -\frac{1}{9} e^{4-3 x}+C $$

Work Step by Step

Given $$\int(x+1) e^{4-3 x} d x $$ Let \begin{align*} u&= x+1\ \ \ \ \ \ dv=e^{4-3 x} d x\\ du&= dx\ \ \ \ \ \ v=\frac{-1}{3}e^{4-3 x} \end{align*} Then \begin{align*} \int(x+1) e^{4-3 x} d x&=\frac{-(x+1)}{3}e^{4-3 x} +\frac{1}{3}\int e^{4-3 x}dx\\ &= \frac{-(x+1)}{3}e^{4-3 x} -\frac{1}{9} e^{4-3 x}+C \end{align*}
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