University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 8 - Practice Exercises - Page 473: 5

Answer

$x^2 e^x+ e^x+c$

Work Step by Step

Integration by parts formula suggests that $\int p'(x) q(x)=p(x) q(x)-\int p(x) q'(x)dx$ $\int (x+1)^2 e^x dx=\int (x^2+2x+1) e^x dx$ or, $\int (x^2+2x+1) e^x dx=\int x^2 e^x dx+\int 2x e^x dx+\int e^x dx=x^2 e^x-\int 2x e^x dx+\int 2x e^x dx+e^x +c$ Thus, we have $x^2 e^x-\int 2x e^x dx+\int 2x e^x dx+e^x +c=x^2 e^x+ e^x+c$
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