Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises - Page 477: 56

Answer

$$\displaystyle{\int x^4e^x\ dx=x^4e^x-4x^{3}e^x+12x^2e^x-24xe^x+24e^x+C}$$

Work Step by Step

$\displaystyle{\int x^ne^x\ dx=x^ne^x-n\int x^{n-1}e^x\ dx}\\ n=4\\ \displaystyle{I=\int x^4e^x\ dx}\\ \displaystyle{I=x^4e^x-4\int x^{3}e^x\ dx}\\ \displaystyle{I=x^4e^x-4\left(x^{3}e^x-3\int x^{2}e^x\ dx\right)}\\ \displaystyle{I=x^4e^x-4x^{3}e^x+12\int x^2e^x\ dx}\\ \displaystyle{I=x^4e^x-4x^{3}e^x+12\left(x^{2}e^x-2\int xe^x\ dx\right)}\\ \displaystyle{I=x^4e^x-4x^{3}e^x+12x^2e^x-24\int xe^x\ dx}\\ \displaystyle{I=x^4e^x-4x^{3}e^x+12x^2e^x-24\left(xe^x-1\int x^0e^x\ dx\right)}\\ \displaystyle{I=x^4e^x-4x^{3}e^x+12x^2e^x-24xe^x+24\int e^x\ dx}\\ \displaystyle{I=x^4e^x-4x^{3}e^x+12x^2e^x-24xe^x+24e^x+C}$
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