Calculus (3rd Edition)

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

Chapter 7 - Exponential Functions - 7.8 Inverse Trigonometric Functions - Exercises - Page 375: 97



Work Step by Step

Let $u=e^x$, then $du=e^x dx$. Now, we have $$\int e^x(e^{2x}+1)^3dx=\int (u^2+1)^3du\\ =\int (u^6+3u^4+3u^3+1 )du\\ =\frac{1}{7}u^7+\frac{3}{5}u^5+\frac{3}{3}u^3+u+c\\ =\frac{1}{7}e^{7x}+\frac{3}{5}e^{5x}+e^{3x}+e^{x}+c$$
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